Chapter 2 Differentiation
1. Derivatives of Some Functions
In class 9, we have studied about the derivatives of algebraic and trignometric functions. We derived the following results.
(1) \(\frac{d}{d x}\left(x^n\right)=n x^{n-1}\)
(2) \(\frac{d}{d x}(\sin x)=\cos x\)
(3) \(\frac{d}{d x}(\cos x)=-\sin x\)
(4) \(\frac{d}{d x}(\tan x)=\sec ^2 x\)
(5) \(\frac{d}{d x}(\cot x)=-{cosec}^2 x\)
(6) \(\frac{d}{d x}(\sec x)=\sec x \tan x\)
(7) \(\frac{d}{d x}({cosec} x)=-{cosec} x \cot x\)
In algebra of derivatives, we have established the following rules.
Some Rules of Differentiation
(1) \(\frac{d}{d x}(u+v)=\left(\frac{d u}{d x}+\frac{d v}{d x}\right)\)
(2) \(\frac{d}{d x}(u-v)=\left(\frac{d u}{d x}-\frac{d v}{d x}\right)\)
(3) \(\frac{d}{d x}(u v)=\left(u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}\right)\)
(4) \(\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\left(v \cdot \frac{d u}{d x}-u \cdot \frac{d v}{d x}\right)}{v^2}\)
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Derivatives of Composite Functions (Chain Rule)
(1) Let y = f(t) and t = g(x). Then, \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right).\)
(2) Let y = f(t), t = g(u) and u = h(x). Then, \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d u} \times \frac{d u}{d x}\right) .\)
This rule may be extended further on more variables.
Solved Examples
Example 1 Differentiate each of the following w.r.t. x:
(1) sinx3
(2) cos3x
(3) tan √x
Solution
(1) Let y = sinx3.
Putting x3 = t, we get
y = sin t and t = x3
⇒ \(\frac{d y}{d t}=\cos t \text { and } \frac{d t}{d x}=3 x^2\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)\)
= (cos t .3x2) = 3x2 cos t = 3x2 cos x3.
Hence, \(\frac{d}{d x}\left(\sin x^3\right)=3 x^2 \cos x^3\).
(2) Let y = cos3x = (cos x)3.
Putting cos x = t, we get
y = t3 and t = cos x
⇒ \(\frac{d y}{d t}=3 t^2 \text { and } \frac{d t}{d x}=-\sin x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)\)
= (-3t2sin x) = (-3 sin x)t2 = (-3sin x cos2x).
Hence, \(\frac{d}{d x}\left(\cos ^3 x\right)=-3 \sin x \cos ^2 x .\)
(3) Let y = tan√x
Putting √x = t, we get
y = tan t and t = √x
⇒ \(\frac{d y}{d t}=\sec ^2 t \text { and } \frac{d t}{d x}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)\)
= \(\left(\sec ^2 t \cdot \frac{1}{2 \sqrt{x}}\right)=\frac{\sec ^2 \sqrt{x}}{2 \sqrt{x}}\) [∵ t = √x]
Hence, \(\frac{d}{d x}(\tan \sqrt{x})=\frac{\sec ^2 \sqrt{x}}{2 \sqrt{x}} .\)
Example 2 Differentiate each of the following w.r.t. x:
(1) (ax + b)m
(2) (2x+3)5
(3) \(\sqrt{a x^2+2 b x+c}\)
Solution
(1) Let y = (ax + b)m.
Putting (ax+b) = t, we get
y = tm and t = (ax+b)
⇒ \(\frac{d y}{d t}=m t^{m-1} \text { and } \frac{d t}{d x}=a\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\left(m t^{m-1} \times a\right)=m a t^{m-1}=m a(a x+b)^{m-1} .\)
∴ \(\frac{d}{d x}(a x+b)^m=m a(a x+b)^{m-1}\)
(2) Let y = (2x+3)5.
Putting (2x+3) = t, we get
y = t5 and t = 2x+3
⇒ \(\frac{d y}{d t}=5 t^4 \text { and } \frac{d t}{d x}=2\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=10 t^4=10(2 x+3)^4 .\)
(3) Let y = \(\sqrt{a x^2+2 b x+c}\)
Putting (ax2 + 2bx + c) = t, we get
y = √t and t = (ax2+2bx+c)
⇒ \(\frac{d y}{d t}=\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}} \text { and } \frac{d t}{d x}=(2 a x+2 b)=2(a x+b)\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{1}{2 \sqrt{t}} \times 2(a x+b)=\frac{(a x+b)}{\sqrt{t}}=\frac{(a x+b)}{\sqrt{a x^2+2 b x+c}} .\)
Example 3 Differentiate sin 3x cos tx w.r.t x.
Solution
Given
sin 3x cos tx w.r.t x.
Let y = sin 3x cos 5x = \(\frac{1}{2}\)[2 cos 5x sin 3x]
= \(\frac{1}{2}\){sin (5x+3x) – sin(5x-3x)}
= \(\frac{1}{2} \cdot(\sin 8 x-\sin 2 x)=\frac{1}{2} \sin 8 x-\frac{1}{2} \sin 2 x .\)
∴ \(\frac{d y}{d x}=\frac{1}{2} \cdot \frac{d}{d x}(\sin 8 x)-\frac{1}{2} \cdot \frac{d}{d x}(\sin 2 x)\)
= \(\left(\frac{1}{2} \cdot 8 \cos 8 x-\frac{1}{2} \times 2 \cos 2 x\right)=(4 \cos 8 x-\cos 2 x) .\)
sin 3x cos tx = \(\left(\frac{1}{2} \cdot 8 \cos 8 x-\frac{1}{2} \times 2 \cos 2 x\right)=(4 \cos 8 x-\cos 2 x) .\)
Example 4 Differentiate sin 2x sin 4x w.r.t. x.
Solution
Given
sin 2x sin 4x w.r.t. x
Let y = sin 2x sin 4x = \(\frac{1}{2}\)(2 sin 4x sin 2x)
= \(\frac{1}{2}\) {cos (4x-2x) – cos (4x+2x)}
= \(\frac{1}{2}\) [cos 2x – cos 6x].
∴ \(\frac{d y}{d x}=\frac{1}{2} \cdot \frac{d}{d x}(\cos 2 x)-\frac{1}{2} \cdot \frac{d}{d x}(\cos 6 x)\)
= \(\frac{1}{2} \cdot(-2 \sin 2 x)-\frac{1}{2} \cdot(-6 \sin 6 x)\)
= (3sin 6x – sin 2x).
sin 2x sin 4x = (3sin 6x – sin 2x).
Example 5 Differentiate \(\sqrt{\frac{1-\tan x}{1+\tan x}}\) w.r.t. x.
Solution
Given
\(\sqrt{\frac{1-\tan x}{1+\tan x}}\) w.r.t. x
Let y = \(\sqrt{\frac{1-\tan x}{1+\tan x}}.\)
Putting \(\frac{(1-\tan x)}{1+\tan x)}=t\), we get
y = √t and t = \(\frac{(1-\tan x)}{1+\tan x)}\).
∴ \(\frac{d y}{d t}=\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}} .\)
And, \(\frac{d t}{d x}=\frac{(1+\tan x) \cdot \frac{d}{d x}(1-\tan x)-(1-\tan x) \cdot \frac{d}{d x}(1+\tan x)}{(1+\tan x)^2}\)
= \(\frac{(1+\tan x)\left(-\sec ^2 x\right)-(1-\tan x)\left(\sec ^2 x\right)}{(1+\tan x)^2}=\frac{-2 \sec ^2 x}{(1+\tan x)^2}\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{1}{2 \sqrt{t}} \times \frac{-2 \sec ^2 x}{(1+\tan x)^2}\)
= \(\frac{-\sec ^2 x}{(1+\tan x)^2} \times \frac{\sqrt{1+\tan x}}{\sqrt{1-\tan x}}\)
= \(\frac{-\sec ^2 x}{(1+\tan x)^{3 / 2}(1-\tan x)^{1 / 2}} .\)
\(\sqrt{\frac{1-\tan x}{1+\tan x}}\) = \(\frac{-\sec ^2 x}{(1+\tan x)^{3 / 2}(1-\tan x)^{1 / 2}} .\)
Example 6 If y = \(\frac{1}{\sqrt{a^2-x^2}} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Given
y = \(\frac{1}{\sqrt{a^2-x^2}} \text {, find } \frac{d y}{d x} \text {. }\)
Putting (a2 – x2) = t, we get
y = \(\frac{1}{\sqrt{t}}=t^{-1 / 2}\) and t = (a2 – x2)
⇒ \(\frac{d y}{d t}=-\frac{1}{2} t^{-3 / 2}=\frac{-1}{2 t^{3 / 2}} \text { and } \frac{d t}{d x}=-2 x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{-1}{2 t^{3 / 2}} \times(-2 x)=\frac{x}{t^{3 / 2}}=\frac{x}{\left(a^2-x^2\right)^{3 / 2}} .\)
Example 7 If y = cos2x2, find \(\frac{d y}{d x}\).
Solution
Given: y = (cosx2)2.
Putting x2 = t and cos t = u, we get
y = u2, u = cos t and t = x2
⇒ \(\frac{d y}{d u}=2 u, \frac{d u}{d t}=-\sin t \text { and } \frac{d t}{d x}=2 x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= (2u) x (-sint) x 2x
= -4 x u sin t = -4x cos t sin t = -4xcosx2sinx2.
∴ \(\frac{d y}{d x}=-4 x \cos x^2 \sin x^2\)
Example 8 If y = sin(cos x2), find \(\frac{d y}{d x}\)
Solution
Given
y = sin(cos x2),
Putting x2 = t and cos x2 = cos t = u, we get
y = sin u, u = cos t and t = x2
⇒ \(\frac{d y}{d u}=\cos u, \frac{d u}{d t}=-\sin t \text { and } \frac{d t}{d x}=2 x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= [cos u x (-sin t) x 2x] = -2x sin t cos u
= -2x sin t cos(cos t) = -2x sin x2 cos(cos x2).
∴ \(\frac{d y}{d x}=-2 x \sin x^2 \cos \left(\cos x^2\right) .\)
Example 9 If y = \(\sin (\sqrt{\sin x+\cos x}), \text { find } \frac{d y}{d x}\)
Solution
Given
y = \(\sin (\sqrt{\sin x+\cos x}), \text { find } \frac{d y}{d x}\)
Putting (sin x + cos x) = t and √t = u, we get
y = sin u, u = √t and t = (sin x + cos x)
⇒ \(\frac{d y}{d u}=\cos u_r \frac{d u}{d t}=\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}} \text { and } \frac{d t}{d x}=(\cos x-\sin x)\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= \(\left\{\cos u \times \frac{1}{2 \sqrt{t}} \times(\cos x-\sin x)\right\}=\frac{\cos \sqrt{t}}{2 \sqrt{t}} \cdot(\cos x-\sin x)\)
= \(\frac{\cos (\sqrt{\sin x+\cos x})}{2 \sqrt{\sin x+\cos x}} \cdot(\cos x-\sin x) \text {. }\)
Example 10 If y = \(\sin [\sqrt{\sin \sqrt{x}}] \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Given
y = \(\sin [\sqrt{\sin \sqrt{x}}] \text {, find } \frac{d y}{d x} \text {. }\)
Putting √x = t, sin√x = sin t = u and \(\sqrt{\sin \sqrt{x}}\) = √u = v, we get
y = sin v, v = √u, u =sin t and t = √x.
∴ \(\frac{d y}{d v}=\cos v ; \frac{d v}{d u}=\frac{1}{2} u^{-1 / 2}=\frac{1}{2 \sqrt{u}} ; \frac{d u}{d t}=\cos t \text { and } \frac{d t}{d x}=\frac{1}{2 \sqrt{x}} \text {. }\)
So, \(\frac{d y}{d x}=\left(\frac{d y}{d v} \times \frac{d v}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)=\left[\cos v \cdot \frac{1}{2 \sqrt{u}} \cdot \cos t \cdot \frac{1}{2 \sqrt{x}}\right]\)
= \(\left[\cos \sqrt{u} \cdot \frac{1}{2 \sqrt{u}} \cos t \cdot \frac{1}{2 \sqrt{x}}\right]\) [∵ v = √u]
= \(\frac{1}{4} \cos (\sqrt{\sin t}) \cdot \frac{1}{\sqrt{\sin t}} \cdot \cos \sqrt{x} \cdot \frac{1}{\sqrt{x}}\) [∵ u = sin t]
= \(\frac{1}{4} \cos (\sqrt{\sin \sqrt{x}}) \cdot \frac{1}{\sqrt{\sin \sqrt{x}}} \cdot \cos \sqrt{x} \cdot \frac{1}{\sqrt{x}}\) [∵ t = √x]
= \(\frac{\cos (\sqrt{\sin \sqrt{x}})}{4 \sqrt{x} \sqrt{\sin \sqrt{x}}} \cdot \cos \sqrt{x}\)
Example 11 If y = \(\frac{5 x}{\sqrt[3]{1-x^2}}+\sin ^2(2 x+3) \text {, find } \frac{d y}{d x}\).
Solution
We have
y = \(5 x\left(1-x^2\right)^{-1 / 3}+\sin ^2(2 x+3)\)
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left\{5 x\left(1-x^2\right)^{-1 / 3}\right\}+\frac{d}{d x}\left\{\sin ^2(2 x+3)\right\}\)
= \(\left\{5 x \cdot\left(\frac{-1}{3}\right)\left(1-x^2\right)^{-13} \cdot(-2 x)+\left(1-x^2\right)^{-13} \cdot 5\right\}+\{2 \sin (2 x+3) \cos (2 x+3) \cdot 2\}\)
= \(\frac{10 x^2}{3\left(1-x^2\right)^{43}}+\frac{5}{\left(1-x^2\right)^{3 / 5}}+2 \sin (4 x+6)\)
= \(\frac{10 x^2+15\left(1-x^2\right)}{3\left(1-x^2\right)^{43}}+2 \sin (4 x+6)\)
= \(\frac{\left(15-5 x^2\right)}{3\left(1-x^2\right)^{4 / 3}}+2 \sin (4 x+6) \text {. }\)
Some Results on Limits
Suppose we have to find \(\lim _{x \rightarrow a} f(x)\), where f(x) contains a series which is capable of being expanded, then after making proper expansion and simplifying, we put x =a.
Some important expansions are given below:
(1) For | x | < 1, we have the bionomial expansion
\((1+x)^n=\left\{1+n x+\frac{n(n-1)}{2 !} x^2+\frac{n(n-1)(n-2)}{3 !} x^3+\ldots\right\}\)(2) \(\left(\frac{x^n-a^n}{x-a}\right)=\left(x^{n-1}+a x^{n-2}+a^2 x^{n-3}+\ldots+a^{n-1}\right)\)
(3) \(e^x=\left\{1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots+\frac{x^n}{n !}+\ldots\right\}\)
(4) \(a^x=\left\{1+x(\log a)+\frac{x^2}{2 !}(\log a)^2+\ldots\right\}\)
(5) \(\log (1+x)=\left\{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\right\}\)
(6) \(\sin x=\left\{x-\frac{x^3}{3 !}+\frac{x^5}{5}-\frac{x^7}{7 !}+\ldots\right\}\)
(7) \(\cos x=\left\{1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+\ldots\right\}\)
(8) \(\tan x=\left\{x+\frac{x^3}{3 !}+\frac{2}{15} x^5+\ldots\right\}\)
We shall also make use of the following theorems on limits. The proof of these theorems is beyond the scope of this book.
Fundamental Theorems on Limits (without proof)
(1) \(\lim _{x \rightarrow a}\{f(x)+g(x)\}=\left\{\lim _{x \rightarrow a} f(x)\right\}+\left\{\lim _{x \rightarrow a} g(x)\right\}\)
(2) \(\lim _{x \rightarrow a}\{f(x)-g(x)\}=\left\{\lim _{x \rightarrow a} f(x)\right\}-\left\{\lim _{x \rightarrow a} g(x)\right\}\)
(3) \(\lim _{x \rightarrow a}\{c \cdot f(x)\}=c \cdot\left\{\lim _{x \rightarrow a} f(x)\right\}\), where c is a constant.
(4) \(\lim _{x \rightarrow a}\{f(x) \cdot g(x)\}=\left\{\lim _{x \rightarrow a} f(x)\right\} \cdot\left\{\lim _{x \rightarrow a} g(x)\right\}\)
(5) \(\lim _{x \rightarrow a}\left\{\frac{f(x)}{g(x)}\right\}=\frac{\left\{\lim _{x \rightarrow a} f(x)\right\}}{\left\{\lim _{x \rightarrow a} g(x)\right\}} \text {, provided } \lim _{x \rightarrow a} g(x) \neq 0 \text {. }\)
(6) If f(x) ≤ g(x) for all x then \(\lim _{x \rightarrow a} f(x) \leq \lim _{x \rightarrow a} g(x) .\)
Some Important Theorems on Limits
Theorem 1 Prove that \(\lim _{x \rightarrow a}\left(\frac{x^n-a^n}{x-a}\right)=n a^{n-1}\), where a > 0.
Proof
We know that \(\left(\frac{x^n-a^n}{x-a}\right)=\left(x^{n-1}+a x^{n-2}+a^2 x^{n-3}+\ldots+a^{n-1}\right) .\)
∴ \(\lim _{x \rightarrow a}\left(\frac{x^n-a^n}{x-a}\right)=\lim _{x \rightarrow a}\left(x^{n-1}+a x^{n-2}+a^2 x^{n-3}+\ldots+a^{n-1}\right)\)
= nan-1 [putting x = a].
Hence, \(\lim _{x \rightarrow a}\left(\frac{x^n-a^n}{x-a}\right)=n a^{n-1} .\)
Theorem 2 Prove that \(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\)
Proof
Expanding log(1+x), we get
\(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots\right)}{x}\)= \(\lim _{x \rightarrow 0}\left(1-\frac{x}{2}+\frac{x^2}{3}-\ldots\right)=1\) [putting x = 0].
Hence, \(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\)
Theorem 3 Prove that \(\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)=1\)
Proof
Expanding ex, we get
\(\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)=\lim _{x \rightarrow 0}\left\{\frac{\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots\right)-1}{x}\right\}\)= \(\lim _{x \rightarrow 0} \frac{\left(x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots\right)}{x}\)
= \(\lim _{x \rightarrow 0}\left(1+\frac{x}{2}+\frac{x^2}{6}+\ldots\right)=1\) [putting x = 0].
Hence, \(\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)=1\)
Theorem 4 Prove that \(\lim _{x \rightarrow 0}(1+x)^{1 / x}=e\)
Proof
Expanding (1+x)1/x we get
\(\lim _{x \rightarrow 0}(1+x)^{1 / x}=\lim _{x \rightarrow 0}\left\{1+\frac{1}{x} \cdot x+\frac{\frac{1}{x}\left(\frac{1}{x}-1\right)}{2 !} \cdot x^2+\frac{\frac{1}{x}\left(\frac{1}{x}-1\right)\left(\frac{1}{x}-2\right)}{3 !} \cdot x^3+\ldots\right\}\)= \(\lim _{x \rightarrow 0}\left\{1+1+\frac{(1-x)}{2 !}+\frac{(1-x)(1-2 x)}{3 !}+\ldots\right\}\)
= \(\left(1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots\right)=e\)
∴ \(\lim _{x \rightarrow 0}(1+x)^{1 / x}=e\)
Theorem 5 Prove that \(\lim _{x \rightarrow-}\left(1+\frac{1}{x}\right)^x=e .\)
Proof
Putting \(\frac{1}{x}\) = y, we get
\(\lim _{x \rightarrow 0}\left(1+\frac{1}{x}\right)^x=\lim _{x \rightarrow 0}(1+y)^{1 / y}=e\) [∵ x → ∞, \(\frac{1}{x}\) → 0 ⇒ y → 0].
Hence, \(\lim _{x \rightarrow-}\left(1+\frac{1}{x}\right)^x=e .\)
Theorem 6 Prove that
(1) \(\lim _{x \rightarrow \infty} \frac{1}{x}=0\) and
(2) \(\lim _{x \rightarrow-\infty} \frac{1}{x}=\infty .\)
Proof
(1) Putting \(\frac{1}{x}\) = y, we get
\(\lim _{x \rightarrow \infty} \frac{1}{x}=\lim _{x \rightarrow \infty} y=0\) [as, x → ∞, \(\frac{1}{x}\) → 0 ⇒ y → 0].
Hence, \(\lim _{x \rightarrow \infty} \frac{1}{x}=0\)
(2) Clearly, when x → 0, then \(\frac{1}{x}\) → ∞.
∴ \(\lim _{x \rightarrow-\infty} \frac{1}{x}=\infty .\)
2. Derivatives of Exponential and Logarithmic Functions
Exponential Function
Let a be a real number such that a > 1.
Then, f(x) = ax is called an exponential function.
Its domain = R and range = R+.
When a = e, we have the exponential function, f(x) = ex.
This is called natural exponential function.
Logarithmic Function
Let a be a real number such that a > 1.
If ax = b, we define, logax = b.
We say that log of x to the base a is equal to b.
log10 x is called common logarithm of x.
logex is called natural logarithm of x and we denote it simply by log x.
It is easy to verify the following results:
(1) log(xy) = (log x) + (log y)
(2) log \frac{x}{y} = (log x) – (log y)
(3) log (xn) = n log x
(4) \(\log _a x=\frac{1}{\log _x a}\)
(5) logxx = 1 and log 1 = 0
Derivatives of Exponential and Logarithmic Functions
We have
(1) \(\frac{d}{d x}\left(e^x\right)=e^x\)
(2) \(\frac{d}{d x}(\log x)=\frac{1}{x}\)
Derivative of ax
Let y = ax. Then, log y = x log a …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\log a\) ⇒ \(\frac{d y}{d x}=y(\log a)\)
⇒ \(\frac{d y}{d x}=a^x(\log a) .\)
Hence, \(\frac{d}{d x}\left(a^x\right)=a^x(\log a) .\)
Summary
(1) \(\frac{d}{d x}\left(e^x\right)=e^x\)
(2) \(\frac{d}{d x}\left(a^x\right)=a^x(\log a)\)
(3) \(\frac{d}{d x}(\log x)=\frac{1}{x}\)
(4) \(\frac{d}{d x}\left(\log _a x\right)=\frac{1}{x \log a}\)
Solved Examples
Example 1 Differentiate each of the following w.r.t. x:
(1) \(e^{x^2}\)
(2) e-3x
(3) ecos x
Solution
(1) Let y = \(e^{x^2}\).
y = et and t = x2
⇒ \(\frac{d y}{d t}=e^t \text { and } \frac{d t}{d x}=2 x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\left(e^t \times 2 x\right)=\left(e^{x^2} \times 2 x\right)=2 x e^{x^2}\)
Hence, \(\frac{d}{d x}\left(e^{x^2}\right)=2 x e^{x^2} .\)
(2) Let y = e-3x.
Puting -3x = t, we get
y = et and t = -3x
⇒ \(\frac{d y}{d t}=e^t \text { and } \frac{d t}{d x}=-3\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=-3 e^t=-3 e^{-3 x} .\)
Hence, \(\frac{d}{d x}\left(e^{-3 x}\right)=-3 e^{-3 x} .\)
(3) Let y = ecos x
Putting cos x = t, we get
y = et and t = cos x
⇒ \(\frac{d y}{d t}=e^t \text { and } \frac{d t}{d x}=-\sin x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\left(-e^t \sin x\right)=\left(-e^{\cos x} \sin x\right)\)
Hence, \(\frac{d}{d x}\left(e^{\cos x}\right)=-e^{\cos x} \sin x .\)
Example 2 Differentiate each of the following w.r.t. x:
(1) sin (log x), x > 0
(2) log(log x), x > 1
Solution
(1) Let y = sin(log x).
Putting log x = t, we get
y = sin t and t = log x
⇒ \(\frac{d y}{d t}=\cos t \text { and } \frac{d t}{d x}=\frac{1}{x}\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\left(\cos t \times \frac{1}{x}\right)=\cos (\log x) \times \frac{1}{x}=\frac{\cos (\log x)}{x} .\)
Hence, \(\frac{d}{d x}\{\sin (\log x)\}=\frac{\cos (\log x)}{x} .\)
(2) Let y = log(log x).
Putting log x = t, we get
y = log t and t = log x
⇒ \(\frac{d y}{d t}=\frac{1}{t} \text { and } \frac{d t}{d x}=\frac{1}{x}\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\left(\frac{1}{t} \times \frac{1}{x}\right)=\left(\frac{1}{\log x} \times \frac{1}{x}\right)=\frac{1}{(x \log x)} .\)
∴ \(\frac{d}{d x}\{\log (\log x)\}=\frac{1}{(x \log x)}\)
Example 3 If y = \(e^{\sqrt{\cot x}} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Given: y = \(e^{\sqrt{\cot x}}\)
Putting cot x = t and \(\sqrt{\cot x}=\sqrt{t}\) = u, we get
y = eu, u = √t and t = cot x
⇒ \(\frac{d y}{d u}=e^u, \frac{d u}{d t}=\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}} \text { and } \frac{d t}{d x}=-{cosec}^2 x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= \(\left\{e^\mu \cdot \frac{1}{2 \sqrt{t}} \cdot\left(-{cosec}^2 x\right)\right\}=e^{\sqrt{\cot x}} \cdot \frac{1}{2 \sqrt{\cot x}} \cdot\left(-{cosec}^2 x\right)\)
= \(\frac{\left(-{cosec}^2 x\right) e^{\sqrt{\cot x}}}{2 \sqrt{\cot x}}\)
Example 4 If y = \(\log \tan \frac{x}{2}, \text { find } \frac{d y}{d x}\)
Solution
Given: y = \(\log \tan \frac{x}{2}\).
Putting \(\frac{x}{2}\) = t and tan\(\frac{x}{2}\) = tan t = u, we get
y = log u, u = tan t and t = \(\frac{x}{2}\)
⇒ \(\frac{d y}{d u}=\frac{1}{2} u^{-t / 2}=\frac{1}{2 \sqrt{u}}, \frac{d u}{d t}=e^t \text { and } \frac{d t}{d x}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= \(\left(\frac{1}{2 \sqrt{u}} \times e^t \times \frac{1}{2 \sqrt{x}}\right)=\left\{\frac{1}{2 \sqrt{u}} \times u \times \frac{1}{2 \sqrt{x}}\right\}=\frac{\sqrt{u}}{4 \sqrt{x}}=\frac{e^{\frac{1}{2} t}}{4 \sqrt{x}}\)
= \(\frac{e^{\frac{1}{2} \sqrt{x}}}{4 \sqrt{x}}\).
Example 5 If y = \(\frac{1}{\log \cos x} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Given: y = (log cos x)-1.
Putting cos x = t and log cos x = log t = u, we get
y = u-1 = \(\frac{1}{u}\), u = log t and t = cos x
⇒ \(\frac{d y}{d u}=\frac{-1}{u^2}, \frac{d u}{d t}=\frac{1}{t} \text { and } \frac{d t}{d x}=-\sin x\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= \(\left\{\frac{-1}{u^2} \times \frac{1}{t} \times(-\sin x)\right\}=\left\{\frac{1}{(\log \cos x)^2} \cdot \frac{1}{\cos x} \cdot \sin x\right\}\)
= \(\frac{\tan x}{(\log \cos x)^2}\).
Example 6 If y = \(\sqrt{e^{\sqrt{x}}} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
y = \(\sqrt{e^{\sqrt{x}}} \text {, find } \frac{d y}{d x} \text {. }\)
Putting √x = t, e√x = et = u, we get
y = √u, u = et and t = √x
⇒ \(\frac{d y}{d u}=\frac{1}{2} u^{-1 / 2}=\frac{1}{2 \sqrt{u}}, \frac{d u}{d t}=e^t \text { and } \frac{d t}{d x}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= \(\left(\frac{1}{2 \sqrt{u}} \times e^t \times \frac{1}{2 \sqrt{x}}\right)=\left\{\frac{1}{2 \sqrt{u}} \times u \times \frac{1}{2 \sqrt{x}}\right\}=\frac{\sqrt{u}}{4 \sqrt{x}}=\frac{e^{\frac{1}{2} t}}{4 \sqrt{x}}\)
= \(\frac{e^{\frac{1}{2} \sqrt{x}}}{4 \sqrt{x}}\)
Example 7 If y = log log log x3, find \(\frac{d y}{d x}\)
Solution
Given
y = log log log x3
Let x3 = t, log x3 = log t = u and log log x3 = log u = v.
Then, y = log v, v = log u, u = log t and t = x3
⇒ \(\frac{d y}{d x}=\left(\frac{d y}{d v} \times \frac{d v}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= \(\left(\frac{1}{v} \times \frac{1}{u} \times \frac{1}{t} \times 3 x^2\right)=\frac{3 x^2}{t u v}\)
= \(\frac{3 x^2}{x^3(\log t)(\log t)}=\frac{3}{x(\log t)(\log \log t)}\)
= \(\frac{3}{x\left(\log x^3\right)\left(\log \log x^3\right)}=\frac{1}{x(\log x)\left(\log \log x^3\right)}\).
Example 8 If y = \(\sqrt{\log \left\{\sin \left(\frac{x^2}{3}-1\right)\right\}} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
y = \(\sqrt{\log \left\{\sin \left(\frac{x^2}{3}-1\right)\right\}} \text {, find } \frac{d y}{d x} \text {. }\)
Put \(\left(\frac{x^2}{3}-1\right)=t, sin \left(\frac{x^2}{3}-1\right)\) = sin t = u
and \(\log \left\{\sin \left(\frac{x^2}{3}-1\right)\right\}=\log u=v\)
Then, y = √v, where v = log u, u = sin t and t = \(\left(\frac{x^2}{3}-1\right)\).
∴ \(\frac{d y}{d v}=\frac{1}{2} v^{-1 / 2}=\frac{1}{2 \sqrt{v}} ; \frac{d v}{d u}=\frac{1}{u} ; \frac{d u}{d t}=\cos t \text { and } \frac{d t}{d x}=\frac{2 x}{3} \text {. }\)
So, \(\frac{d y}{d x}=\left(\frac{d y}{d v} \times \frac{d v}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)\)
= \(\left(\frac{1}{2 \sqrt{v}} \cdot \frac{1}{u} \cdot \cos t \cdot \frac{2 x}{3}\right)\)
= \(\frac{x}{3} \cdot \frac{\cos t}{u \cdot \sqrt{\log u}}=\frac{x}{3} \cdot \frac{\cos t}{\sin t \sqrt{\log \sin t}}\)
[∵ v = log u and u = sin t]
= \(\frac{x \cot t}{3 \sqrt{\log \sin t}}=\frac{x \cot \left(\frac{x^2}{3}-1\right)}{3 \cdot \sqrt{\log \sin \left(\frac{x^2}{3}-1\right)}}\) [∵ t = \(\left(\frac{x^2}{3}-1\right)\)].
Example 9 If y = ex log (sin 2x), find \(\frac{d y}{d x}\).
Solution
y = ex log (sin 2x)
We have
\(\frac{d y}{d x}=\frac{d}{d x}\left\{e^x \log (\sin 2 x)\right\}\)= \(e^x \cdot \frac{d}{d x}\{\log (\sin 2 x)\}+\log (\sin 2 x) \cdot \frac{d}{d x}\left(e^x\right)\)
= \(e^x \cdot\left\{\frac{1}{\sin 2 x} \cdot \cos 2 x \cdot 2\right\}+\log (\sin 2 x) \cdot e^x\)
= 2 ex cot 2x + ex log(sin 2x)
= ex {2 cot 2x + log(sin 2x)}.
Example 10 If y = eax cos(bx+c), find \(\frac{d y}{d x}\).
Solution
y = eax cos(bx+c)
We have
\(\frac{d y}{d x}=e^{a x} \cdot \frac{d}{d x}\{\cos (b x+c)\}+\cos (b x+c) \cdot \frac{d}{d x}\left(e^{a x}\right)\)= eax . {-b sin(bx+c)} + cos(bx+c) . aeax
= eax. {a cos (bx+c) – b sin(bx+c)}.
Example 11 Differentiate \(\log \sqrt{\frac{1+\cos ^2 x}{\left(1-e^{2 x}\right)}}\)
Solution
\(\log \sqrt{\frac{1+\cos ^2 x}{\left(1-e^{2 x}\right)}}\)Let y = \(\log \sqrt{\frac{1+\cos ^2 x}{\left(1-e^{2 x}\right)}}=\log \left(\frac{1+\cos ^2 x}{1-e^{2 x}}\right)^{1 / 2}=\frac{1}{2} \log \left(\frac{1+\cos ^2 x}{1-e^{2 x}}\right)\)
∴ \(y=\frac{1}{2} \log \left(1+\cos ^2 x\right)-\frac{1}{2} \log \left(1-e^{2 x}\right)\)
⇒ \(\frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\left(1+\cos ^2 x\right)}(2 \cos x)(-\sin x)-\frac{1}{2} \cdot \frac{1}{\left(1-e^{2 x}\right)} \cdot\left(-2 e^{2 x}\right)\)
= \(\left\{\frac{-\sin x \cos x}{\left(1+\cos ^2 x\right)}+\frac{e^{2 x}}{\left(1-e^{2 x}\right)}\right\} .\)
Hence, \(\frac{d}{d x}\left\{\log \sqrt{\frac{1+\cos ^2 x}{\left(1-e^{2 x}\right)}}\right\}=\left\{\frac{-\sin x \cos x}{\left(1+\cos ^2 x\right)}+\frac{e^{2 x}}{\left(1-e^{2 x}\right)}\right\} \text {. }\)
Example 12 If y = \(\log \sqrt{\frac{1+\sin ^2 x}{1-\sin x}}, \text { find } \frac{d y}{d x}\).
Solution
We have
y = \(\frac{1}{2} [log (1+sin2x) – log(1-sin x)]\) …(1)
On differentiating (1) w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{2} \cdot\left[\frac{d}{d x}\left\{\log \left(1+\sin ^2 x\right)\right\}-\frac{d}{d x}\{\log (1-\sin x)\}\right]\)= \(\frac{1}{2} \cdot\left\{\frac{2 \sin x \cos x}{\left(1+\sin ^2 x\right)}-\frac{(-\cos x)}{(1-\sin x)}\right\}\)
= \(\frac{1}{2} \cdot\left\{\frac{\sin 2 x}{\left(1+\sin ^2 x\right)}+\frac{\cos x}{(1-\sin x)}\right\} .\)
Example 13 If y = log sin(ex + 5x + 8), find \(\frac{d y}{d x}\).
Solution
Given: y = log sin(ex + 5x + 8) …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{\sin \left(e^x+5 x+8\right)} \cdot \cos \left(e^x+5 x+8\right) \cdot \frac{d}{d x}\left(e^x+5 x+8\right)\)= {cot (ex + x + 8)} (ex + 5) = (ex + 5). cot (ex + 5x + 8).
Example 14 If y = \(\sqrt{x^2+1}-\log \left\{\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\right\} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
We have
y = \(\sqrt{x^2+1}-\log \left\{\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\right\}\)
⇒ \(y=\sqrt{x^2+1}-\log \left\{1+\sqrt{x^2+1}\right\}+\log x\)
⇒ \(\frac{d y}{d x}=\frac{1}{2}\left(x^2+1\right)^{-1 / 2} \cdot 2 x-\frac{1}{\left\{1+\sqrt{x^2+1}\right\}} \cdot\left\{\frac{1}{2}\left(x^2+1\right)^{-1 / 2} \cdot 2 x\right\}+\frac{1}{x}\)
= \(\frac{x}{\sqrt{x^2+1}}-\frac{1}{\left\{1+\sqrt{x^2+1}\right\}} \cdot \frac{x}{\sqrt{x^2+1}}+\frac{1}{x}\)
= \(\frac{x\left\{1+\sqrt{x^2+1}\right\}-x}{\left(\sqrt{x^2+1}\right)\left\{1+\sqrt{x^2+1}\right\}}+\frac{1}{x}=\frac{x \sqrt{x^2+1}}{\left(\sqrt{x^2+1}\right)\left\{1+\sqrt{x^2+1}\right\}}+\frac{1}{x}\)
= \(\frac{x}{\left\{1+\sqrt{x^2+1}\right\}}+\frac{1}{x}=\frac{\left(x^2+1\right)+\sqrt{x^2+1}}{x\left\{1+\sqrt{x^2+1}\right\}}\)
= \(\frac{\left(\sqrt{x^2+1}\right)\left\{\left(\sqrt{x^2+1}\right)+1\right\}}{x\left\{1+\sqrt{x^2+1}\right\}}=\frac{\sqrt{x^2+1}}{x} .\)
3. Derivaties of Inverse Trigonometric Functions
In the table given below, we mention the domain and range of various inverse trigonometric functions.
Derivatives of Inverse Trigonometric Functions
Example 1 Prove that \(\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}\), where x ∈ ]-1, 1[.
Solution
\(\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}\), where x ∈ ]-1, 1[
Let y = sin-1x, where x ∈ ]-1,1[ and y ∈ \(]-\frac{\pi}{2}, \frac{\pi}{2}[\). Then,
y = sin-1x ⇒ x = sin y
⇒ \(\frac{d x}{d y}=\cos y \geq 0\) since y = \(]-\frac{\pi}{2}, \frac{\pi}{2}[\)
⇒ \(\frac{d x}{d y}=\sqrt{1-\sin ^2 y}=\sqrt{1-x^2}\)
⇒ \(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}} .\)
Example 2 Prove that \(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}\), where x ∈ ]-1, 1[.
Solution
\(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}\), where x ∈ ]-1, 1[.
Let y = cos-1x, where x ∈ ]-1, 1[ and y ∈ \(]0, \frac{\pi}{2}[\). Then,
y = cos-1x ⇒ x = cos y
⇒ \(\frac{d x}{d y}=-\sin y\), where sin y > 0, since y ∈ \(]0, \frac{\pi}{2}[\)
⇒ \(\frac{d x}{d y}=-\sqrt{1-\cos ^2 y}=-\sqrt{1-x^2}\)
⇒ \(\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^2}}\)
Hence, \(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}} .\)
Example 3 Prove that \(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{\left(1+x^2\right)}\), where x ∈ R.
Solution
\(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{\left(1+x^2\right)}\), where x ∈ R
Let y = tan-1x, where x ∈ R and y ∈ \(]-\frac{\pi}{2}, \frac{\pi}{2}[\). Then,
x = tan y
⇒ \(\frac{d x}{d y}=\sec ^2 y=\left(1+\tan ^2 y\right)=\left(1+x^2\right)\)
⇒ \(\frac{d y}{d x}=\frac{1}{\left(1+x^2\right)} .\)
Hence, \(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{\left(1+x^2\right)} .\)
Example 4 Prove that \(\frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{\left(1+x^2\right)}\), where x ∈ R.
Solution
\(\frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{\left(1+x^2\right)}\), where x ∈ R
Let y = cot-1x, where x ∈ R and y = ]0,π[. Then,
x = cot y
⇒ \(\frac{d x}{d y}=-{cosec}^2 y=-\left(1+\cot ^2 y\right)=-\left(1+x^2\right)\)
⇒ \(\frac{d y}{d x}=\frac{-1}{\left(1+x^2\right)} .\)
Hence, \(\frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{\left(1+x^2\right)}\).
Example 5 Prove that \(\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}}\), where x ∈ R – [-1,1].
Solution
\(\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}}\), where x ∈ R – [-1,1]
Let y = sec-1x, where x ∈ R – [-1,1] and y ∈ ]0,π[ – {\(\frac{\pi}{2}\)}. Then,
x = sec y
⇒ \(\frac{d x}{d y}=\sec y \tan y>0\)
⇒ \(\frac{d y}{d x}=\frac{1}{\sec y \tan y}=\frac{1}{\sec y \cdot \sqrt{\sec ^2 y-1}}\)
⇒ \(\frac{d y}{d x}=\frac{1}{|x| \sqrt{x^2-1}}\)
Hence, \(\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}} .\)
Example 6 Prove that \(\frac{d}{d x}\left({cosec}^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2-1}}\), where x ∈ R – [-1,1].
Solution
\(\frac{d}{d x}\left({cosec}^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2-1}}\), where x ∈ R – [-1,1]
Let y = cosex-1x, where x ∈ R – [-1,1] and y ∈ \(]-\frac{\pi}{2}, \frac{\pi}{2}[\) – {0}. Then,
x = cosec y
⇒ \(\frac{d x}{d y}= -{cosec} y \cot y, \text { where } {cosec} y \cot y>0\)
⇒ \(\frac{d y}{d x}=\frac{-1}{{cosec} y \cot y}=\frac{-1}{({cosec} y) \sqrt{{cosec}^2 y-1}}=\frac{-1}{|x| \sqrt{x^2-1}} .\)
Summary
(1) \(\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}\)
(2) \(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}\)
(3) \(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{\left(1+x^2\right)}\)
(4) \(\frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{\left(1+x^2\right)}\)
(5) \(\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}}\)
(6) \(\frac{d}{d x}\left({cosec}^{-1} x\right)=\frac{-1}{|x| \cdot \sqrt{x^2-1}}\)
Solved Examples
Example 1 Differentiate the following w.r.t. x:
(1) sin-12x
(2) tan-1√x
(3) cos-1(cot x)
Solution
(1) Let y = sin-12x.
Putting 2x = t, we get y = sin-1t and t = 2x.
Now, y = sin-1t ⇒ \(\frac{d y}{d t}=\frac{1}{\sqrt{1-t^2}}\)
And, t = 2x ⇒ \(\frac{d t}{d x}=2\).
∴ \(\frac{d u}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{2}{\sqrt{1-t^2}}=\frac{2}{\sqrt{1-4 x^2}}\) [∵ t = 2x].
Hence, \(\frac{d}{d x}\left(\sin ^{-1} 2 x\right)=\frac{2}{\sqrt{1-4 x^2}} .\)
(2) Let y = tan-1√x.
Putting √x = t, we get y = tan-1t and t = √x.
Now, y = tan-1t ⇒ \(\frac{d y}{d t}=\frac{1}{\left(1+t^2\right)}\)
And, t = √x ⇒ \(\frac{d t}{d x}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}}\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{1}{\left(1+t^2\right)} \cdot \frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{x}(1+x)}\) [∵ t = √x]
Hence, \(\frac{d}{d x}\left(\tan ^{-1} \sqrt{x}\right)=\frac{1}{2 \sqrt{x}(1+x)} .\)
(3) Let y = cos-1(cot x).
Putting cot x = t, we get y = cos-1t and t = cot x.
Now, y = cos-1t ⇒ \(\frac{d y}{d t}=\frac{-1}{\sqrt{1-t^2}}\).
And, t = cot x ⇒ \(\frac{d t}{d x}=-{cosec}^2 x .\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{{cosec}^2 x}{\sqrt{1-t^2}}=\frac{{cosec}^2 x}{\sqrt{1-\cot ^2 x}}\) [∵ t = cot x].
Hence, \(\frac{d}{d x}\left\{\cos ^{-1}(\cot x)\right\}=\frac{{cosec}^2 x}{\sqrt{1-\cot ^2 x}}\)
Example 2 Differentiate the following w.r.t. x:
(1) sec(tan-1x)
(2) sin(tan-1x)
(3) cot(cos-1x)
Solution
(1) Let y = sec(tan-1x).
Putting tan-1x = t, we get y = sec t and t = tan-1x.
Now, y = sec t ⇒ \(\frac{d y}{d t}=\sec t \tan t\).
And, t = tan-1x ⇒ \(\frac{d t}{d x}=\frac{1}{\left(1+x^2\right)}\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{\sec t \tan t}{\left(1+x^2\right)}=\frac{\left(\sqrt{1+\tan ^2 t}\right)(\tan t)}{\left(1+x^2\right)}\)
= \(\frac{\left(\sqrt{1+x^2}\right) x}{\left(1+x^2\right)}=\frac{x}{\sqrt{1+x^2}}\) [∵ t = tan-1x ⇒ tan t = x].
Hence, \(\frac{d}{d x}\left\{\sec \left(\tan ^{-1} x\right)\right\}=\frac{x}{\sqrt{1+x^2}}\)
(2) Let y = sin(tan-1x).
Putting tan-1x = t, we get y = sin t and t = tan-1x.
Now, y = sin t ⇒ \(\frac{d y}{d t}=\cos t .\)
And, t = tan-1x ⇒ \(\frac{d t}{d x}=\frac{1}{\left(1+x^2\right)}\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\cos t \cdot \frac{1}{\left(1+x^2\right)}=\frac{1}{\left(1+x^2\right)^{3 / 2}}\)
[∵ tan t = x ⇒ \(\cos t=\frac{1}{\sqrt{1+x^2}}\)].
Hence, \(\frac{d}{d x}\left\{\sin \left(\tan ^{-1} x\right)\right\}=\frac{1}{\left(1+x^2\right)^{3 / 2}}\)
(3) Let y = cot(cos-1x).
Putting cos-1x = t, we get y = cot t and t = cos-1x.
Now, y = cot t ⇒ \(\frac{d y}{d t}=-{cosec}^2 t\).
And, t = cos-1x ⇒ \(\frac{d t}{d x}=\frac{-1}{\sqrt{1-x^2}} .\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{{cosec}^2 t}{\sqrt{1-x^2}}=\frac{1}{\left(1-x^2\right)^{3 / 2}}\)
[∵ cos t = x ⇒ \({cosec}^2 t=\frac{1}{\left(1-x^2\right)}\)].
Hence, \(\frac{d}{d x}\left\{\cot \left(\cos ^{-1} x\right)\right\}=\frac{1}{\left(1-x^2\right)^{3 / 2}}\)
Example 3 If y = sin(tan-12x), prove that \(\frac{d y}{d x}=\frac{2}{\left(1+4 x^2\right)^{3 / 2}} \text {. }\).
Solution
y = sin(tan-12x)
Putting tan-12x = t, we get y = sin t and t = tan-12x.
Now, y = sin t ⇒ \(\frac{d y}{d t}=\cos t .\)
And, t = tan-12x ⇒ \(\frac{d t}{d x}=\left\{\frac{1}{\left(1+4 x^2\right)} \times 2\right\}=\frac{2}{\left(1+4 x^2\right)} \text {. }\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\left\{\cos t \times \frac{2}{\left(1+4 x^2\right)}\right\}\) …(1)
Now, t = tan-12x ⇒ tan t = 2x
⇒ \(\sec t=\sqrt{1+\tan ^2 t}=\sqrt{1+4 x^2}\)
⇒ \(\cos t=\frac{1}{\sec t}=\frac{1}{\sqrt{1+4 x^2}}\) …(2)
Putting the value of cos t from (2) in (1), we get
\(\frac{d y}{d x}=\left\{\frac{1}{\sqrt{1+4 x^2}} \times \frac{2}{\left(1+4 x^2\right)}\right\}=\frac{2}{\left(1+4 x^2\right)^{3 / 2}} .\)Example 4 Differentiate \(\sqrt{\cot ^{-1} \sqrt{x}}\) w.r.t. x.
Solution
\(\sqrt{\cot ^{-1} \sqrt{x}}\) w.r.t. x
Let y = \(\sqrt{\cot ^{-1} \sqrt{x}}\).
Putting √x = t and cot-1√x = cot-1t = u, we get
y = √u, where u = cot-1t and t = √x.
Now, y = √u ⇒ \(\frac{d y}{d u}=\frac{1}{2} u^{-1 / 2}=\frac{1}{2 \sqrt{u}}\)
u = cot-1t ⇒ \(\frac{d u}{d t}=\frac{-1}{\left(1+t^2\right)}\).
And, t = √x ⇒ \(\frac{d t}{d x}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}} .\)
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)=\frac{-1}{4 \sqrt{u}\left(1+t^2\right) \sqrt{x}}\)
= \(\frac{-1}{4\left(\sqrt{\cot ^{-1} t}\right)\left(1+t^2\right) \sqrt{x}}\) [∵ u = cot-1t]
= \(\frac{-1}{4\left(\sqrt{\cot ^{-1} \sqrt{x}}\right)(1+x) \sqrt{x}}\) [∵ t = √x].
Example 5 Differentiate etan-1√x w.r.t. x.
Solution
etan-1√x w.r.t. x
Let y = etan-1√x.
Putting √x = t and tan-1√x = tan-1t = u, we get
y = eu, where u = tan-1t and t = √x.
Now, y = eu ⇒ \(\frac{d y}{d u}=e^u\);
u = tan-1t ⇒ \(\frac{d u}{d t}=\frac{1}{\left(1+t^2\right)}\)
And, t = √x ⇒ \(\frac{d t}{d x}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}}\).
∴ \(\frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)=e^x \cdot \frac{1}{\left(1+t^2\right)} \cdot \frac{1}{2 \sqrt{x}}\)
= \(e^{\tan ^{-1} t} \cdot \frac{1}{\left(1+t^2\right)} \cdot \frac{1}{2 \sqrt{x}}\) [∵ u = tan-1t]
= \(\frac{e^{\tan ^{-1} \sqrt{x}}}{2 \sqrt{x}(1+x)}\) [∵ t = √x].
Hence, \(\frac{d y}{d x}=\frac{e^{\tan ^{-1} \sqrt{x}}}{2 \sqrt{x}(1+x)}\)
Example 6 If y = \(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
y = \(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}} \text {, find } \frac{d y}{d x} \text {. }\)
y = \(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}\) …(1)
⇒ \(y \sqrt{1-x^2}=x \sin ^{-1} x \) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(y \cdot \frac{d}{d x}\left(\sqrt{1-x^2}\right)+\left(\sqrt{1-x^2}\right) \frac{d y}{d x}=x \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)+\sin ^{-1} x \cdot \frac{d}{d x}(x)\)⇒ \(y \cdot \frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}} \cdot(-2 x)+\left(\sqrt{1-x^2}\right) \frac{d y}{d x}=x \cdot \frac{1}{\sqrt{1-x^2}}+\sin ^{-1} x \cdot 1\)
⇒ \(\frac{-x y}{\sqrt{1-x^2}}+\left(\sqrt{1-x^2}\right) \frac{d y}{d x}=\frac{x}{\sqrt{1-x^2}}+\sin ^{-1} x\)
⇒ \(\frac{-x y}{\sqrt{1-x^2}}+\left(\sqrt{1-x^2}\right) \frac{d y}{d x}=\frac{x}{\sqrt{1-x^2}}+\sin ^{-1} x\) [using(1)]
⇒ \(-x^2 \sin ^{-1} x+\left(1-x^2\right)^{3 / 2} \frac{d y}{d x}=x\left(\sqrt{1-x^2}\right)+\left(1-x^2\right) \sin ^{-1} x\)
⇒ \(\left(1-x^2\right)^{3 / 2} \frac{d y}{d x}=x\left(\sqrt{1-x^2}\right)+\sin ^{-1} x\)
⇒ \(\frac{d y}{d x}=\frac{x\left(\sqrt{1-x^2}\right)+\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} .\)
Example 7 Find \(\frac{d}{d x}\left[\left(\sqrt{1-x^2}\right) \sin ^{-1} x-x\right]\)
Solution
We have
\(\frac{d}{d x}\left[\left(\sqrt{1-x^2}\right) \sin ^{-1} x-x\right]\)= \(\frac{d}{d x}\left[\left(\sqrt{1-x^2}\right) \sin ^{-1} x\right]-\frac{d}{d x}(x)\)
= \(\left(\sqrt{1-x^2}\right) \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)+\left(\sin ^{-1} x\right) \cdot \frac{d}{d x}\left(\sqrt{1-x^2}\right)-1\)
= \(\left(\sqrt{1-x^2}\right) \cdot \frac{1}{\left(\sqrt{1-x^2}\right)}+\left(\sin ^{-1} x\right) \cdot \frac{1}{2}\left(1-x^2\right)^{-1 / 2} \cdot(-2 x)-1\)
= \(\left\{1-\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}-1\right\}=\frac{-x \sin ^{-1} x}{\sqrt{1-x^2}} \text {. }\)
Example 8 Show that \(\frac{d}{d x}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]=\sqrt{a^2-x^2} .\)
Solution
We have
\(\frac{d}{d x}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]\)= \(\frac{d}{d x}\left[\frac{x}{2} \cdot \sqrt{a^2-x^2}\right]+\frac{a^2}{2} \cdot \frac{d}{d x}\left[\sin ^{-1} \frac{x}{a}\right]\)
= \(\frac{x}{2} \cdot \frac{d}{d x}\left(\sqrt{a^2-x^2}\right)+\left(\sqrt{a^2-x^2}\right) \cdot \frac{d}{d x}\left(\frac{x}{2}\right)+\frac{a^2}{2} \cdot \frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \cdot \frac{1}{a}\)
= \(\frac{x}{2} \cdot \frac{1}{2}\left(a^2-x^2\right)^{\frac{-1}{2}} \cdot(-2 x)+\left(\sqrt{a^2-x^2}\right) \cdot \frac{1}{2}+\frac{a^2}{2 \sqrt{a^2-x^2}}\)
= \(\frac{-x^2}{2 \sqrt{a^2-x^2}}+\frac{\sqrt{a^2-x^2}}{2}+\frac{a^2}{2 \sqrt{a^2-x^2}}\)
= \(\frac{-x^2+\left(a^2-x^2\right)+a^2}{2 \sqrt{a^2-x^2}}=\frac{\left(a^2-x^2\right)}{\sqrt{\left(a^2-x^2\right)}}=\sqrt{\left(a^2-x^2\right)}\)
Hence, \(\frac{d}{d x}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]=\sqrt{a^2-x^2} .\)
4. Differentiation by Trigonometrical Transformations
Some Useful Results
(1) \((1-\cos x)=2 \sin ^2\left(\frac{x}{2}\right)\)
(2) \((1+\cos x)=2 \cos ^2\left(\frac{x}{2}\right)\)
(3) Sin 3x = (3 sin x – 4 sin3x)
(4) cos 3x = (4 cos3x – 3 cos x)
(5) \(\sin x=\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}\)
(6) \(\cos x=\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)}\)
(7) \(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left\{\frac{x-y}{1+x y}\right\}\)
(8) \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left\{\frac{x+y}{1-x y}\right\}\)
Some Useful Substitutions
Suppose we are given sin-1f(x), cos-1f(x), tan-1f(x), etc.
Rule 1. If f(x) = \(\sqrt{a^2-x^2}\), put x = a sin θ or x = a cos θ.
Rule 2. If f(x) = \(\sqrt{a^2+x^2}\), put x = a tan θ or x = a cot θ.
Rule 3. If f(x) = \(\sqrt{x^2-a^2}\), put x = a sec θ or x = a cosec θ.
Rule 4. If f(x) = \(\sqrt{a-x}\), put x = a cos 2θ.
Solved Examples
Example 1 Differentiate each of the following w.r.t. x:
(1) \(\tan ^{-1}\left(\frac{1-\cos x}{\sin x}\right)\)
(2) \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\)
Solution
(1) Let y = \(\tan ^{-1}\left(\frac{1-\cos x}{\sin x}\right)=\tan ^{-1}\left\{\frac{2 \sin ^2(x / 2)}{2 \sin (x / 2) \cos (x / 2)}\right\}\)
= \(\tan ^{-1}\left\{\tan \frac{x}{2}\right\}=\frac{x}{2}\)
∴ y = \(\frac{x}{2}\).
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2} .\)
(2) Let y = \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)\)
[on dividing num. and denom. by cos x]
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-x\right)\right\}=\left(\frac{\pi}{4}-x\right)\)
∴ \(y=\left(\frac{\pi}{4}-x\right)\)
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}-x\right)=\frac{d}{d x}\left(\frac{\pi}{4}\right)-\frac{d}{d x}(x)=(0-1)=-1\).
Example 2 Differentiate w.r.t. x:
(1) \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\tan ^{-1}\left\{\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}\right\}\)
(2) tan-1(sec x + tan x)
Solution
(1) Let y = \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\tan ^{-1}\left\{\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}\right\}\)
= \(\tan ^{-1}\left\{\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right\}\)
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
∴ \(y=\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{d}{d x}\left(\frac{\pi}{4}\right)-\frac{d}{d x}\left(\frac{x}{2}\right)=\left(0-\frac{1}{2}\right)=\frac{-1}{2} .\)
(2) Let y = tan-1(sec x + tan x)
= \(\tan ^{-1}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)\)
= \(\tan ^{-1}\left\{\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{\sin \left(\frac{\pi}{2}+x\right)}\right\}\)
{∵ \(\cos \left(\frac{\pi}{2}+x\right)=-\sin x ; \sin \left(\frac{\pi}{2}+x\right)=\cos x\)}
= \(\tan ^{-1}\left\{\frac{2 \sin ^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}\right\}\)
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}=\left(\frac{\pi}{4}+\frac{x}{2}\right) \text {. }\)
∴ \(y=\left(\frac{\pi}{4}+\frac{x}{2}\right)\)
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{d}{d x}\left(\frac{x}{2}\right)=\left(0+\frac{1}{2}\right)=\frac{1}{2} .\)
Example 3 Differentiate w.r.t x:
(1) \(\tan ^{-1}\left\{\sqrt{\frac{1+\cos x}{1-\cos x}}\right\}\)
(2) \(\tan ^{-1}\left\{\sqrt{\frac{1+\sin x}{1-\sin x}}\right\}\)
Solution
(1) Let y = \(\tan ^{-1}\left\{\sqrt{\frac{1+\cos x}{1-\cos x}}\right\}=\tan ^{-1}\left\{\sqrt{\frac{2 \cos ^2(x / 2)}{2 \sin ^2(x / 2)}}\right\}\)
= \(\tan ^{-1}\left(\cot \frac{x}{2}\right)=\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right\}=\left(\frac{\pi}{2}-\frac{x}{2}\right)\)
∴ \(y=\left(\frac{\pi}{2}-\frac{x}{2}\right)\).
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{2}\right)=\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{d}{d x}\left(\frac{x}{2}\right)=\left(0-\frac{1}{2}\right)=\frac{-1}{2} .\)
(2) Let y = \(\tan ^{-1}\left\{\sqrt{\frac{1+\sin x}{1-\sin x}}\right\}=\tan ^{-1}\left\{\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{1+\cos \left(\frac{\pi}{2}+x\right)}\right\}^{\frac{1}{2}}\)
= \(\tan ^{-1}\left\{\frac{2 \sin ^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}\right\}^{\frac{1}{2}}=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}=\left(\frac{\pi}{4}+\frac{x}{2}\right) .\)
∴ \(y=\left(\frac{\pi}{4}+\frac{x}{2}\right)\)
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{d}{d x}\left(\frac{x}{2}\right)=\left(0+\frac{1}{2}\right)=\frac{1}{2} .\)
Example 4 Differentiate \(\cos ^{-1}\left\{\sqrt{\frac{1+\cos x}{2}}\right\}\) w.r.t. x.
Solution
Let y = \(\cos ^{-1}\left\{\sqrt{\frac{1+\cos x}{2}}\right\}=\cos ^{-1}\left\{\sqrt{\frac{2 \cos ^2(x / 2)}{2}}\right\}\)
= \(\cos ^{-1}\{\cos (x / 2)\}=\frac{x}{2}\)
∴ y = \(\frac{x}{2}\)
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2} .\)
Example 5 If y = \(\cot ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}, \text { find } \frac{d y}{d x}\).
Solution
We have
\(y=\cot ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}=\cot ^{-1} \sqrt{\frac{1+\cos \left(\frac{\pi}{2}+x\right)}{1-\cos \left(\frac{\pi}{2}+x\right)}}\)= \(\cot ^{-1} \sqrt{\frac{2 \cos ^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \sin ^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}}=\cot ^{-1}\left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}=\left(\frac{\pi}{4}+\frac{x}{2}\right) \text {. }\)
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{d}{d x}\left(\frac{x}{2}\right)=\left(0+\frac{1}{2}\right)=\frac{1}{2} .\)
Example 6 If y = \(\cot ^{-1} \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \text { and } 0<x<\frac{\pi}{2} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
We have
\((1+\sin x)=\left\{\cos ^2(x / 2)+\sin ^2(x / 2)+2 \sin (x / 2) \cos (x / 2)\right\}\)= \(\{\cos (x / 2)+\sin (x / 2)\}^2 .\)
\((1-\sin x)=\left\{\cos ^2(\pi / 2)+\sin ^2(\pi / 2)-2 \sin (\pi / 2) \cos (\pi / 2)\right\}\)= \(\{\cos (x / 2)-\sin (x / 2)\}^2 .\)
∴ \(\sqrt{1+\sin x}=\sqrt{\{\cos (x / 2)+\sin (x / 2)\}^2}=\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\)
and \(\sqrt{1-\sin x}=\sqrt{\{\cos (x / 2)-\sin (x / 2)\}^2}=\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\)
∴ y = \(\cot ^{-1}\left\{\frac{[\cos (x / 2)+\sin (x / 2)]+[\cos (x / 2)-\sin (x / 2)]}{[\cos (x / 2)+\sin (x / 2)]-[\cos (x / 2)-\sin (x / 2)]}\right\}\)
= \(\cot ^1\left\{\frac{2 \cos (x / 2)}{2 \sin (x / 2)}\right\}=\cot ^{-1}\{\cot (x / 2)\}=\frac{x}{2}\)
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2}\)
Example 7 If y = \(\tan ^{-1} \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
We have
y = \(\tan ^{-1}\left\{\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})} \times \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\right\}\)
= \(\tan ^{-1}\left\{\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{1-\sin ^2 x}}{(1+\sin x)-(1-\sin x)}\right\}=\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\)
= \(\tan ^{-1}\left\{\frac{2 \cos ^2(x / 2)}{2 \sin (x / 2) \cos (x / 2)}\right\}=\tan ^{-1}\left\{\cot \frac{x}{2}\right\}=\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right\}\)
= \(\left(\frac{\pi}{2}-\frac{x}{2}\right)\)
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{2}\right)=\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{d}{d x}\left(\frac{x}{2}\right)=\left(0-\frac{1}{2}\right)=-\frac{1}{2} .\)
Example 8 Differentiate w.r.t. x:
(1) \(\cot ^{-1}\left(\frac{1}{x}\right)\)
(2) \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
(3) \(\cot ^{-1}\left(\frac{1-x}{1+x}\right)\)
Solution
(1) Let y = \(\cot ^{-1}\left(\frac{1}{x}\right)\)
Putting x = tan θ, we get
\(y=\cot ^{-1}\left(\frac{1}{\tan \theta}\right)=\cot ^{-1}(\cot \theta)=\theta=\tan ^{-1} x \text {. }\)∴ \(\frac{d y}{d x}=\frac{1}{\left(1+x^2\right)}\).
(2) Let y = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
Putting x = tan θ, we get
\(y=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)=\tan ^{-1}(\tan 2 \theta)=2 \theta=2 \tan ^{-1} x\).
∴ \(\frac{d y}{d x}=\frac{2}{\left(1+x^2\right)}\)
Hence, \(\frac{d}{d x}\left\{\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right\}=\frac{2}{\left(1+x^2\right)} .\)
(3) Let y = \(\cot ^{-1}\left(\frac{1-x}{1+x}\right)\)
Putting x = tan θ, we get
\(y=\cot ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cot ^{-1}\left\{\tan \left(\frac{\pi}{4}-\theta\right)\right\}\)= \(\cot ^{-1}\left[\cot \left\{\frac{\pi}{2}-\left(\frac{\pi}{4}-\theta\right)\right\}\right]=\left(\frac{\pi}{4}+\theta\right)=\frac{\pi}{4}+\tan ^{-1} x.\)
∴ \(\frac{d y}{d x}=\frac{1}{\left(1+x^2\right)}\).
Hence, \(\frac{d}{d x}\left\{\cot ^{-1}\left(\frac{1-x}{1+x}\right)\right\}=\frac{1}{\left(1+x^2\right)} .\)
Example 9 Differentiate w.r.t. x:
(1) \(\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
(2) \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
(3) \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)\)
Solution
(1) Let y = \(\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
Putting x = tan θ, we get
\(y=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)=\cos ^{-1}(\cos 2 \theta)=2 \theta=2 \tan ^{-1} x \text {. }\)∴ \(\frac{d y}{d x}=\frac{2}{\left(1+x^2\right)}.\)
Hence, \(\frac{d}{d x}\left\{\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right\}=\frac{2}{\left(1+x^2\right)}.\)
(2) Let y = \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Putting x = tan θ, we get
\(y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta=2 \tan ^{-1} x \text {. }\)∴ \(\frac{d y}{d x}=\frac{2}{\left(1+x^2\right)} .\)
Hence, \(\frac{d}{d x}\left\{\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right\}=\frac{2}{\left(1+x^2\right)} .\)
(3) Let y = \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)\)
Putting x = cos θ, we get
\(y=\sec ^{-1}\left(\frac{1}{2 \cos ^2 \theta-1}\right)=\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right)\)= sec-1(sec 2θ) = 2θ = 2 cos-1x.
∴ y = 2 cos-1x.
Hence, \(\frac{d y}{d x}=2 \frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-2}{\sqrt{1-x^2}} .\)
Example 10 Differentiate w.r.t. x:
(1) cos-1(4x3 – 3x)
(2) \(\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
(3) \(\sec ^{-1}\left(\frac{x^2+1}{x^2-1}\right)\)
Solution
(1) Let y = cos-1(4x3 – 3x).
Putting x = cos θ, we get
y = cos-1(4 cos3θ – 3 cos θ) = cos-1(cos 3 θ) = 3θ.
∴ y = 3θ ⇒ y = 3 cos-1x
⇒ \(\frac{d y}{d x}=\frac{-3}{\sqrt{1-x^2}}\)
(2) Let y = \(\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
Putting x = tan θ, we get
\(y=\sin ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\cos 2 \theta)=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-2 \theta\right)\right]\)= \(\left(\frac{\pi}{2}-2 \theta\right)=\left(\frac{\pi}{2}-2 \tan ^{-1} x\right)\)
∴ y = \(\left(\frac{\pi}{2}-2 \tan ^{-1} x\right)\).
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-2 \tan ^{-1} x\right)=\frac{d}{d x}\left(\frac{\pi}{2}\right)-2 \cdot \frac{d}{d x}\left(\tan ^{-1} x\right)\)
= \(\left\{0-\frac{2}{\left(1+x^2\right)}\right\}=\frac{-2}{\left(1+x^2\right)}\).
(3) Let y = \(\sec ^{-1}\left(\frac{x^2+1}{x^2-1}\right)\)
Putting x = cot θ, we get
y = \(\sec ^{-1}\left(\frac{\cot ^2 \theta+1}{\cot ^2 \theta-1}\right)=\sec ^{-1}\left(\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)\)
= sec-1 (sec 2θ) = 2θ = 2 cot-1x.
∴ \(\frac{d y}{d x}=\frac{-2}{\left(1+x^2\right)}\).
Hence, \(\frac{d}{d x}\left\{\sec ^{-1}\left(\frac{x^2+1}{x^2-1}\right)\right\}=\frac{-2}{\left(1+x^2\right)}\).
Example 11 Differentiate \(\cot ^{-1}\left(\frac{1-x}{1+x}\right)\) w.r.t. x.
Solution
Let y = \(\cot ^{-1}\left(\frac{1-x}{1+x}\right)\)
Putting x = tan θ, we get
y = \(\cot ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cot ^{-1}\left\{\tan \left(\frac{\pi}{4}-\theta\right)\right\}\)
= \(\cot ^{-1}\left[\cot \left\{\frac{\pi}{2}-\left(\frac{\pi}{4}-\theta\right)\right]\right]=\cot ^{-1}\left[\cot \left(\frac{\pi}{4}+\theta\right)\right]\)
= \(\left(\frac{\pi}{4}+\theta\right)\)
= \(\frac{\pi}{4}+\tan ^{-1} x\) [∵ x = tan θ ⇒ θ = tan-1x].
∴ \(y=\frac{\pi}{4}+\tan ^{-1} x\)
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left\{\frac{\pi}{4}+\tan ^{-1} x\right\}=\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{d}{d x}\left(\tan ^{-1} x\right)\)
= \(\left\{0+\frac{1}{\left(1+x^2\right)}\right\}=\frac{1}{\left(1+x^2\right)}\).
Hence, \(\frac{d y}{d x}=\frac{1}{\left(1+x^2\right)}\).
Example 12 Differentiate \(\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-1}{x}\right\}\) w.r.t x.
Solution
Let y = \(\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-1}{x}\right\}\).
Putting x = tan θ, we get
y = \(\tan ^{-1}\left\{\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right\}=\tan ^{-1}\left\{\frac{\sec \theta-1}{\tan \theta}\right\}\)
= \(\tan ^{-1}\left\{\frac{\left(\frac{1}{\cos \theta}-1\right)}{\sin \theta} \cdot \cos \theta\right\}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= \(\tan ^{-1}\left\{\frac{2 \sin ^2(\theta / 2)}{2 \sin (\theta / 2) \cos (\theta / 2)}\right\}=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\}\)
= \(\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x\)
∴ y = \(\frac{1}{2}\) tan-1x
⇒ \(\frac{d y}{d x}=\frac{1}{2} \cdot \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{2\left(1+x^2\right)}\)
Example 13 If \(y=\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right\} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Putting x2 = cos θ, we get
y = \(\tan ^{-1}\left\{\frac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}\right\}\)
= \(\tan ^{-1}\left\{\frac{\sqrt{2 \cos ^2(\theta / 2)}-\sqrt{2 \sin ^2(\theta / 2)}}{\sqrt{2 \cos ^2(\theta / 2)}+\sqrt{2 \sin ^2(\theta / 2)}}\right\}\)
= \(\tan ^{-1}\left\{\frac{\sqrt{2} \cos (\theta / 2)-\sqrt{2} \sin (\theta / 2)}{\sqrt{2} \cos (\theta / 2)+\sqrt{2} \sin (\theta / 2)}\right\}=\tan ^{-1}\left\{\frac{\cos (\theta / 2)-\sin (\theta / 2)}{\cos (\theta / 2)-\sin (\theta / 2)}\right\}\)
= \(\tan ^{-1}\left\{\frac{1-\tan (\theta / 2)}{1+\tan (\theta / 2)}\right\}=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right\}\)
= \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x^2\)
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left\{\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x^2\right\}=\frac{d}{d x}\left(\frac{\pi}{4}\right)-\frac{1}{2} \cdot \frac{d}{d x}\left(\cos ^{-1} x^2\right)\)
= \(\left\{0-\frac{1}{2} \cdot \frac{-1}{\sqrt{1-x^4}} \cdot 2 x\right\}=\frac{x}{\sqrt{1-x^4}}\)
Example 14 If y = \(\sin ^{-1}\left[x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right] \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Putting x = sin θ and √x = sin Φ, we get
y = sin-1[sin θ cos Φ – sin Φ cos θ] = sin-1[sin (θ-Φ)]
= (θ – Φ) = sin-1x – sin-1√x.
∴ y = sin-1x – sin-1√x
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left\{\sin ^{-1} x-\sin ^{-1} \sqrt{x}\right\}\)
= \(\frac{d}{d x}\left(\sin ^{-1} x\right)-\frac{d}{d x}\left\{\sin ^{-1} \sqrt{x}\right\}\)
= \(\left[\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x} \cdot \sqrt{1-x}}\right]\)
Example 15 Differentiate \(\tan ^{-1}\left(\frac{x^{1 / 3}+a^{1 / 3}}{1-x^{1 / 3} a^{1 / 3}}\right)\) w.r.t. x.
Solution
Let y = \(\tan ^{-1}\left(\frac{x^{1 / 3}+a^{1 / 3}}{1-x^{1 / 3} a^{1 / 3}}\right)\).
Putting x1/3 = tan θ and a1/3 = tan Φ, we get
y = \(\tan ^{-1}\left(\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}\right)=\tan ^{-1}[\tan (\theta+\phi)]\)
= (θ + Φ) = tan-1(x1/3) + tan-1(a1/3).
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left\{\tan ^{-1}\left(x^{1 / 3}\right)+\tan ^{-1}\left(a^{1 / 3}\right)\right\}\)
= \(\frac{d}{d x}\left\{\tan ^{-1}\left(x^{1 / 3}\right)\right\}+\frac{d}{d x}\left\{\tan ^{-1}\left(a^{1 / 3}\right)\right\}=\frac{1}{\left(1+x^{2 / 3}\right)} \cdot \frac{1}{3} x^{-2 / 3}\)
= \(\frac{1}{3 x^{2 / 3}\left(1+x^{2 / 3}\right)}\) [∵ tan-1(a1/3) = constant].
Example 16 Differentiate \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)\) w.r.t. x.
Solution
Let y = \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)\).
Putting a = r sin θ and b = r cos θ, we get
y = \(\tan ^{-1}\left\{\frac{r(\sin \theta \cos x-\cos \theta \sin x)}{r(\cos \theta \cos x+\sin \theta \sin x)}\right\}\)
= \(\tan ^{-1}\left\{\frac{\sin (\theta-x)}{\cos (\theta-x)}\right\}=\tan ^{-1}\{\tan (\theta-x)\}\)
= \(\theta-x=\left(\tan ^{-1} \frac{a}{b}-x\right)\) [∵ \(\frac{a}{b}\) = tan θ ⇒ θ = tan-1 \(\frac{a}{b}\)].
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1} \frac{a}{b}-x\right)\)
= \(\frac{d}{d x}\left(\tan ^{-1} \frac{a}{b}\right)-\frac{d}{d x}(x)=-1\) [∵ tan-1 \(\frac{a}{b}\) = constant].
Example 17 If y = \(\sin ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right\} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Putting x = cos 2θ, we get
y = \(\sin ^{-1}\left\{\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{2}\right\}\)
= \(\sin ^{-1}\left\{\frac{\sqrt{2 \cos ^2 \theta}-\sqrt{2 \sin ^2 \theta}}{2}\right\}=\sin ^{-1}\left\{\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{2}\right\}\)
= \(\sin ^{-1}\left\{\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta\right\}=\sin ^{-1}\left\{\sin \frac{\pi}{4} \cos \theta-\cos \frac{\pi}{4} \sin \theta\right\}\)
= \(\sin ^{-1}\left\{\sin \left(\frac{\pi}{4}-\theta\right)\right\}\)
= \(\left(\frac{\pi}{4}-\theta\right)=\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right)\)
[∵ x = cos 2θ ⇒ 2θ = cos-1x ⇒ θ = \(\frac{1}{2}\)cos-1x]
∴ y = \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)
Hence, \(\frac{d y}{d x}=\frac{d}{d x}\left\{\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right\}=\frac{d}{d x}\left(\frac{\pi}{4}\right)-\frac{1}{2} \frac{d}{d x}\left(\cos ^{-1} x\right)\)
= \(\left\{0-\frac{1}{2} \cdot \frac{(-1)}{\sqrt{1-x^2}}\right\}=\frac{1}{2 \sqrt{1-x^2}}\).
Example 18 Differentiate each of the following w.r.t. x:
(1) \(\tan ^{-1}\left(\sqrt{1+x^2}+x\right)\)
(2) \(\tan ^{-1}\left(\sqrt{1+x^2}-x\right)\)
Solution
(1) Let y = \(\tan ^{-1}\left(\sqrt{1+x^2}+x\right)\)
Putting x = cot θ, we get
y = \(\tan ^{-1}({cosec} \theta+\cot \theta)=\tan ^{-1}\left(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
= \(\tan ^{-1}\left(\frac{1+\cos \theta}{\sin \theta}\right)=\tan ^{-1}\left(\frac{2 \cos ^2(\theta / 2)}{2 \sin (\theta / 2) \cos (\theta / 2)}\right)\)
= \(\tan ^{-1}\left(\cot \frac{\theta}{2}\right)=\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}\)
= \(\frac{\pi}{2}-\frac{1}{2} \theta=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x .\)
∴ \(\frac{d y}{d x}=-\frac{1}{2} \cdot \frac{-1}{\left(1+x^2\right)}=\frac{1}{2\left(1+x^2\right)}\)
Hence, \(\frac{d}{d x}\left\{\tan ^{-1}\left(\sqrt{1+x^2}+x\right)\right\}=\frac{1}{2\left(1+x^2\right)} .\)
(2) Let y = \(\tan ^{-1}\left(\sqrt{1+x^2}-x\right)\)
Putting x = cot θ, we get
y = \(\tan ^{-1}({cosec} \theta-\cot \theta)=\tan ^{-1}\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)\)
= \(\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1}\left\{\frac{2 \sin ^2(\theta / 2)}{2 \sin (\theta / 2) \cos (\theta / 2)}\right\}\)
= \(\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{1}{2} \theta=\frac{1}{2} \cot ^{-1} x\)
∴ \(\frac{d y}{d x}=\frac{-1}{2\left(1+x^2\right)}\).
Hence, \(\frac{d}{d x}\left\{\tan ^{-1}\left(\sqrt{1+x^2}-x\right)\right\}=\frac{-1}{2\left(1+x^2\right)}\).
Example 19 Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right)\) w.r.t. x.
Solution
Let y = \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right)\).
Putting x = tan θ, we get
y = \(\tan ^{-1}\left(\frac{\sec \theta+1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1+\cos \theta}{\sin \theta}\right)\)
= \(\tan ^{-1}\left\{\frac{2 \cos ^2(\theta / 2)}{2 \sin (\theta / 2) \cos (\theta / 2)}\right\}=\tan ^{-1}\left\{\cot \frac{\theta}{2}\right\}\)
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}=\left(\frac{\pi}{2}-\frac{\theta}{2}\right)=\frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x\)
∴ \(\frac{d y}{d x}=-\frac{1}{2\left(1+x^2\right)} .\)
Hence, \(\frac{d}{d x}\left\{\tan ^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right)\right\}=\frac{-1}{2\left(1+x^2\right)}\).
Example 20 Differentiate \(\cot ^{-1}\left(\sqrt{1+x^2}+x\right)\) w.r.t. x.
Solution
Let y = \(\cot ^{-1}\left(\sqrt{1+x^2}+x\right)\).
Putting x = cot θ, we get
y = \(\cot ^{-1}({cosec} \theta+\cot \theta)=\cot ^{-1}\left(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
= \(\cot ^{-1}\left(\frac{1+\cos \theta}{\sin \theta}\right)=\cot ^{-1}\left\{\frac{2 \cos ^2(\theta / 2)}{2 \sin (\theta / 2) \cos (\theta / 2)}\right\}\)
= \(\cot ^{-1}\left(\cot \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2} \cot ^{-1} x \text {. }\)
∴ \(\frac{d y}{d x}=\frac{-1}{2\left(1+x^2\right)}\).
Hence, \(\frac{d}{d x}\left\{\cot ^{-1}\left(\sqrt{1+x^2}+x\right)\right\}=\frac{-1}{2\left(1+x^2\right)}\).
Example 21 Differentiate \(\cos ^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)\) w.r.t. x.
Solution
Let y = \(\cos ^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)=\cos ^{-1}\left\{\frac{\left(x-\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)}\right\}=\cos ^{-1}\left(\frac{x^2-1}{x^2+1}\right) \text {. }\)
Putting x = tan θ, we get
y = \(\cos ^{-1}\left(\frac{\tan ^2 \theta-1}{\tan ^2 \theta+1}\right)=\cos ^{-1}(-\cos 2 \theta)=\cos ^{-1}\{\cos (\pi-2 \theta)\}\)
= (π – 2θ) = π – 2 tan-1x.
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left(\pi-2 \tan ^{-1} x\right)=\frac{d}{d x}(\pi)-2 \cdot \frac{d}{d x}\left(\tan ^{-1} x\right)=\left(0-\frac{2}{1+x^2}\right)\)
= \(\frac{-2}{\left(1+x^2\right)}\).
Example 22 If y = \(\tan ^{-1}\left(\frac{\sqrt{a}-\sqrt{x}}{1+\sqrt{a x}}\right) \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Putting √a = tan α and √x = tan θ, we get
y = \(\tan ^{-1}\left(\frac{\tan \alpha-\tan \theta}{1+\tan \alpha \tan \theta}\right)=\tan ^{-1}\{\tan (\alpha-\theta)\}=(\alpha-\theta)\)
∴ y = (α – θ) ⇒ y = tan-1√a – tan-1√x
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1} \sqrt{a}\right)-\frac{d}{d x}\left(\tan ^{-1} \sqrt{x}\right)\)
= \(\left\{0-\frac{1}{(1+x)} \cdot \frac{1}{2} x^{-1 / 2}\right\}=\frac{-1}{2 \sqrt{x}(1+x)}\).
Hence, \(\frac{d y}{d x}=\frac{-1}{2 \sqrt{x}(1+x)}\).
Example 23 If y = \(\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^2}}{13}\right\}, \text { find } \frac{d y}{d x} \text {. }\)
Solution
We have
y = \(\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^2}}{13}\right\}\)
Let \(\frac{5}{13}\) = sin α and x = cos θ. Then,
cos α = \(\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}\)
and \(\sqrt{1-x^2}=\sqrt{1-\cos ^2 \theta}=\sqrt{\sin ^2 \theta}=\sin \theta\)
∴ y = sin-1{sin α cos θ + cos α sin θ}
= sin-1{sin(α+θ)}
= \(\alpha+\theta=\sin ^{-1} \frac{5}{13}+\cos ^{-1} x .\)
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left\{\sin ^{-1} \frac{5}{13}+\cos ^{-1} x\right\}=\frac{d}{d x}\left\{\sin ^{-1} \frac{5}{13}\right\}+\frac{d}{d x}\left(\cos ^{-1} x\right)\)
= \(\left\{0-\frac{1}{\sqrt{1-x^2}}\right\}=\frac{-1}{\sqrt{1-x^2}}\)
Hence, \(\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^2}}\)
5. Differentiation of Implicit Function
Let f(x,y) = a be a function of x and y defined in such a manner that y is not expressible directly in terms of x. Then, f(x,y) = a is called an implicit function of x and y. In differentiating such a function, we differentiate both sides of the equation termwise, keeping in mind that
\(\frac{d}{d x}\left(y^2\right)=2 y \cdot \frac{d y}{d x} ; \frac{d}{d x}\left(y^3\right)=3 y^2 \cdot \frac{d y}{d x},\) and so on.
Solved Examples
Example 1 If x3 + y3 = 3axy, find \(\frac{d x}{d y}\).
Solution
Given: x3 + y3 = 3axy …(1)
Differentiating both sides of (1) w.r.t. x, we get
\(3 x^2+3 y^2 \cdot \frac{d y}{d x}=3 a \cdot\left\{x \cdot \frac{d y}{d x}+y \cdot 1\right\}\)⇒ \(3\left(y^2-a x\right) \cdot \frac{d y}{d x}=3\left(a y-x^2\right)\)
⇒ \(\frac{d y}{d x}=\left(\frac{a y-x^2}{y^2-a x}\right) .\)
Example 2 If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, find \(\frac{d x}{d y}\).
Solution
Given: ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 …(1)
Differentiating both sides of (1) w.r.t. x, we get
\(2 a x+2 h\left(x \cdot \frac{d y}{d x}+y \cdot 1\right)+2 b y \cdot \frac{d y}{d x}+2 g+2 f \cdot \frac{d y}{d x}=0\)⇒ \((2 a x+2 h y+2 g)+(2 h x+2 b y+2 f) \cdot \frac{d y}{d x}=0\)
⇒ \(\frac{d y}{d x}=-\left(\frac{a x+h y+g}{h x+b y+f}\right) .\)
Example 3 If \(\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y) \text {, prove that } \frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}} \text {. }\)
Solution
Given: \(\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)\) …(1)
Putting x = sin θ and y = sin Φ, it becomes
cos θ + cos Φ = a(sin θ – sin Φ)
⇒ \(\frac{\cos \theta+\cos \phi}{\sin \theta-\sin \phi}=a\)
⇒ \(\frac{2 \cos \left(\frac{\theta+\phi}{2}\right) \cos \left(\frac{\theta-\phi}{2}\right)}{2 \cos \left(\frac{\theta+\phi}{2}\right) \sin \left(\frac{\theta-\phi}{2}\right)}=a\)
⇒ \(\cot \left(\frac{\theta-\phi}{2}\right)=a \Rightarrow \theta-\phi=2 \cot ^{-1} a\)
⇒ sin-1x – sin-1y = 2 cot-1a …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}} \cdot \frac{d y}{d x}=0 .\)Hence, \(\frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}} .\)
Example 4 If \(x \sqrt{1-y^2}+y \sqrt{1-x^2}=1, \text { prove that } \frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}} .\)
Solution
We have \(x \sqrt{1-y^2}+y \sqrt{1-x^2}=1\) …(1)
Putting x = sin θ and y = cos Φ in (1), we get
sin θ cos Φ + cos θ sin Φ = 1
⇒ sin(θ + Φ)
⇒ (θ + Φ) = sin-1(1)
⇒ sin-1x + sin-1y = \(\frac{\pi}{2}\) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}} \cdot \frac{d y}{d x}=0\)∴ \(\frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\)
Example 5 If \(x \sqrt{1+y}+y \sqrt{1+x}=0 \text {, prove that } \frac{d y}{d x}=\frac{-1}{(1+x)^2} \text {. }\)
Solution
\(x \sqrt{1+y}+y \sqrt{1+x}=0\)⇒ \(x \sqrt{1+y}=-y \sqrt{1+x}\)
⇒ x2(1+y) = y2(1+x) [on squaring both sides]
⇒ (x-y)(x + y + xy) = 0
⇒ x + y +xy = 0 [∵ x = y does not satisfy the given equation]
⇒ \(y=\frac{-x}{1+x} \text {. }\)
∴ \(\frac{d y}{d x}=-\left\{\frac{(1+x) \cdot 1-x \cdot 1}{(1+x)^2}\right\}=\frac{-1}{(1+x)^2}\)
Example 6 If sin y = x sin (a+y), prove that \(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a} .\)
Solution
sin y = x sin (a+y)
⇒ \(x=\frac{\sin y}{\sin (a+y)}\) …(1)
On differentiating both sides of (1) w.r.t. y, we get
\(\frac{d x}{d y}=\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin ^2(a+y)}\) [using the quotient rule]
= \(\frac{\sin (a+y-y)}{\sin ^2(a+y)}=\frac{\sin a}{\sin ^2(a+y)}\)
Hence, \(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a} .\)
Example 7 If \(y \cdot \sqrt{x^2+1}=\log \left(\sqrt{x^2+1}-x\right)\), show that \(\left(x^2+1\right) \frac{d y}{d x}+x y+1=0 .\)
Solution
Given: \(y \cdot \sqrt{x^2+1}=\log \left(\sqrt{x^2+1}-x\right)\) …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(y \cdot \frac{1}{2}\left(x^2+1\right)^{-1 / 2} \cdot 2 x+\left(\sqrt{x^2+1}\right) \cdot \frac{d y}{d x}\)= \(\frac{1}{\left\{\sqrt{x^2+1}-x\right\}} \cdot\left\{\frac{1}{2}\left(x^2+1\right)^{-1 / 2} \cdot 2 x-1\right\}\)
⇒ \(\frac{x y}{\sqrt{x^2+1}}+\left(\sqrt{x^2+1}\right) \frac{d y}{d x}=\frac{1}{\left\{\sqrt{x^2+1}-x\right\}} \cdot\left\{\frac{x}{\sqrt{x^2+1}}-1\right\}\)
⇒ \(\frac{x y}{\sqrt{x^2+1}}+\left(\sqrt{x^2+1}\right) \frac{d y}{d x}=\frac{1}{\left\{\sqrt{x^2+1}-x\right\}} \cdot \frac{\left\{x-\sqrt{x^2+1}\right\}}{\sqrt{x^2+1}}\)
⇒ \(x y+\left(x^2+1\right) \frac{d y}{d x}=-1\)
⇒ \(\left(x^2+1\right) \frac{d y}{d x}+x y+1=0 .\)
Example 8 If \(y=\sqrt{x}+\frac{1}{\sqrt{x}}\), show that [/latex]2 x \frac{d y}{d x}+y=2 \sqrt{x} .[/latex]
Solution
\(y=\sqrt{x}+\frac{1}{\sqrt{x}}\) ⇒ √x y = x + 1 …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\sqrt{x} \cdot \frac{d y}{d x}+y \cdot \frac{1}{2} x^{-1 / 2}=1\)⇒ \(\sqrt{x} \frac{d y}{d x}+\frac{y}{2 \sqrt{x}}=1\)
⇒ \(2 x \frac{d y}{d x}+y=2 \sqrt{x}\)
Example 9 If cos(x+y) = y sin x, find \(\frac{d x}{d y}\).
Solution
Given : cos(x+y) = y sin x …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(-\sin (x+y) \cdot \frac{d}{d x}(x+y)=y \cos x+\sin x \cdot \frac{d y}{d x}\)⇒ \(-\sin (x+y) \cdot\left(1+\frac{d y}{d x}\right)=y \cos x+\sin x \cdot \frac{d y}{d x}\)
⇒ \(\{\sin (x+y)+\sin x\} \cdot \frac{d y}{d x}=-\{\sin (x+y)+y \cos x\}\)
⇒ \(\frac{d y}{d x}=\frac{-\{\sin (x+y)+y \cos x\}}{\{\sin (x+y)+\sin x\}} .\)
Example 10 Find \(\frac{d x}{d y}\) when \(\sin (x y)+\frac{x}{y}=x^2-y\)
Solution
Given: \(\sin (x y)+\frac{x}{y}=x^2-y\)
Differentiating both sides w.r.t. x, we get
\(\cos (x y) \cdot \frac{d}{d x}(x y)+x \cdot\left(-\frac{1}{y^2}\right) \frac{d y}{d x}+\frac{1}{y} \cdot 1=2 x-\frac{d y}{d x}\)⇒ \(\cos (x y) \cdot\left[x \cdot \frac{d y}{d x}+y \cdot 1\right]-\frac{x}{y^2} \cdot \frac{d y}{d x}+\frac{1}{y}=2 x-\frac{d y}{d x}\)
⇒ \(\left[x \cos (x y)-\frac{x}{y^2}+1\right] \cdot \frac{d y}{d x}=2 x-\frac{1}{y}-y \cos (x y)\)
⇒ \(\left\{x y^2 \cos (x y)-x+y^2\right\} \cdot \frac{d y}{d x}=2 x y^2-y-y^3 \cos (x y) .\)
Hence, \(\frac{d y}{d x}=\left\{\frac{2 x y^2-y-y^3 \cos (x y)}{x y^2 \cos (x y)-x+y^2}\right\} .\)
Example 11 If ex + ey = ex+y, prove that \(\frac{d x}{d y}\) = -ey-x.
Solution
Given: ex + ey = ex+y …(1)
On dividing through out by ex+y, we get
e-y + e-x = 1 …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(e^{-y} \cdot\left(\frac{-d y}{d x}\right)+e^{-x}(-1)=0\)⇒ \(\frac{d y}{d x}=\frac{-e^{-x}}{e^{-y}}=-e^{(y-x)} .\)
Example 12 If \(\tan ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=a, \text { show that } \frac{d y}{d x}=\frac{x(1-\tan a)}{y(1+\tan a)}\)
Solution
\(\tan ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=a ⇒ \frac{\left(x^2-y^2\right)}{\left(x^2+y^2\right)}=\tan a\)∴ (x2 – y2) = (x2 + y2) tan a …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(2 x-2 y \cdot \frac{d y}{d x}=2 x \tan a+2 y \cdot \frac{d y}{d x} \cdot \tan a\)⇒ \(y(1+\tan a) \frac{d y}{d x}=x(1-\tan a)\)
⇒ \(\frac{d y}{d x}=\frac{x(1-\tan a)}{y(1+\tan a)}\)
6. Differentiation Using Logarithms
When the given function is a power of some expression or a product of expressions, then we take the logarithm on both sides and differentiate termwise, as shown below.
Solved Examples
Example 1 If y = \(\sqrt{\frac{(x-3)\left(x^2+4\right)}{\left(3 x^2+4 x+5\right)}} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Given: y = \(\sqrt{\frac{(x-3)\left(x^2+4\right)}{\left(3 x^2+4 x+5\right)}}\) …(1)
Taking logarithm on both sides of (1), we get
log y = \(\frac{1}{2}\) {log(x-3) + log(x2 + 4) – log(3x2 + 4x + 5)}.
Differentiating both sides w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2} \cdot\left\{\frac{1}{(x-3)}+\frac{2 x}{\left(x^2+4\right)}-\frac{(6 x+4)}{\left(3 x^2+4 x+5\right)}\right\}\)⇒ \(\frac{d y}{d x}=\left(\frac{1}{2} y\right) \cdot\left\{\frac{1}{(x-3)}+\frac{2 x}{\left(x^2+4\right)}-\frac{(6 x+4)}{\left(3 x^2+4 x+5\right)}\right\}\)
= \(\frac{1}{2} \cdot \sqrt{\frac{(x-3)\left(x^2+4\right)}{\left(3 x^2+4 x+5\right)}} \cdot\left\{\frac{1}{(x-3)}+\frac{2 x}{\left(x^2+4\right)}-\frac{(6 x+4)}{\left(3 x^2+4 x+5\right)}\right\} \text {. }\)
Example 2 If y = \(\frac{\sqrt{x}(x+4)^{3 / 2}}{(4 x-3)^{43}} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
Given: \(y \frac{\sqrt{x}(x+4)^{3 / 2}}{(4 x-3)^{43}}\) …(1)
Taking logarithm on both sides of (1), we get
log y = \(\frac{1}{2}\) log x + \(\frac{3}{2}\) log (x+4) – \(\frac{4}{3}\) log (4x – 3).
On differentiating both sides w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{x}+\frac{3}{2} \cdot \frac{1}{(x+4)}-\frac{4}{3} \cdot \frac{4}{(4 x-3)}\)⇒ \(\frac{d y}{d x}=y\left[\frac{1}{2 x}+\frac{3}{2(x+4)}-\frac{16}{3(4 x-3)}\right]\)
= \(\frac{\sqrt{x}(x+4)^{3 / 2}}{(4 x-3)^{4 / 3}} \cdot\left[\frac{1}{2 x}+\frac{3}{2(x+4)}-\frac{16}{3(4 x-3)}\right]\)
Example 3 Differentiate (x+1)2(x+2)3(x+3)4 w.r.t. x.
Solution
(x+1)2(x+2)3(x+3)4 w.r.t. x
Let y = (x+1)2(X+2)3(x+3)4 …(1)
Taking logarithm on both sides of (1), we get
log y = 2 log(x+1) + 3 log(x+2) + 4log(x+3) …(2)
Differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\frac{2}{(x+1)}+\frac{3}{(x+2)}+\frac{4}{(x+3)}\)⇒ \(\frac{d y}{d x}=y \cdot\left[\frac{2}{(x+1)}+\frac{3}{(x+2)}+\frac{4}{(x+3)}\right]\)
= \((x+1)^2(x+2)^3(x+3)^4 \cdot\left[\frac{2}{(x+1)}+\frac{3}{(x+2)}+\frac{4}{(x+3)}\right]\)
Example 4 Differentiate \(\sqrt{(x-1)(x-2)(x-3)(x-4)}\) w.r.t. x.
Solution
\(\sqrt{(x-1)(x-2)(x-3)(x-4)}\) w.r.t. x
Let y = \(\sqrt{(x-1)(x-2)(x-3)(x-4)}\) …(1)
Taking logarithm on both sides of (1), we get
log y = \(\frac{1}{2}\) {log (x-1) + log (x+2) + log (x-3) + log (x-4)} …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2} \cdot\left\{\frac{1}{(x-1)}+\frac{1}{(x-2)}+\frac{1}{(x-3)}+\frac{1}{(x-4)}\right\}\)⇒ \(\frac{d y}{d x}=\left(\frac{y}{2}\right) \cdot\left\{\frac{1}{(x-1)}+\frac{1}{(x-2)}+\frac{1}{(x-3)}+\frac{1}{(x-4)}\right\}\)
= \(\frac{1}{2} \cdot \sqrt{(x-1)(x-2)(x-3)(x-4)}\left\{\frac{1}{(x-1)}+\frac{1}{(x-2)}+\frac{1}{(x-3)}+\frac{1}{(x-4)}\right\} .\)
Example 5 Differentiate (excos3xsin2x) w.r.t x.
Solution
(excos3xsin2x) w.r.t x
Let y = excos3xsin2x …(1)
Taking logarithm on both sides of (1), we get
log y = x + 3 log cos x + 2 log sin x …(2)
\(\frac{1}{y} \cdot \frac{d y}{d x}=1+\frac{3}{\cos x} \cdot(-\sin x)+\frac{2}{\sin x} \cdot \cos x\)⇒ \(\frac{d y}{d x}=y \cdot\{1-3 \tan x+2 \cot x\}\)
= (excos3xsin2x)(1-3 tan x + 2 cot x).
Example 6 Differentiate (tan x tan 2x tan 3x tan 4x) w.r.t. x.
Solution
(tan x tan 2x tan 3x tan 4x) w.r.t. x
Let y = tan x tan 2x tan 3x tan 4x …(1)
Taking logarithm on both sides of (1), we get
log y = log tan x + log tan 2x + log tan 3x + log tan 4x …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\left\{\frac{\sec ^2 x}{\tan x}+\frac{2 \sec ^2 2 x}{\tan 2 x}+\frac{3 \sec ^2 3 x}{\tan 3 x}+\frac{4 \sec ^2 4 x}{\tan 4 x}\right\}\)⇒ \(\frac{d y}{d x}=y \cdot\left[\frac{1}{\sin x \cos x}+\frac{2}{\sin 2 x \cos 2 x}\right.\left.+\frac{3}{\sin 3 x \cos 3 x}+\frac{4}{\sin 4 x \cos 4 x}\right]\)
= \(y \cdot\left[\frac{2}{\sin 2 x}+\frac{4}{\sin 4 x}+\frac{6}{\sin 6 x}+\frac{8}{\sin 8 x}\right]\)
= [2 tan x tan 2x tan 3x tan 4x] x [coses 2x + 2 cosec 4x + 3 cosec 6x + 4 cosec 8x].
Example 7 If y = (2x + 3)(3x – 5), find \(\frac{d x}{d y}\).
Solution
y = (2x + 3)(3x – 5)
Given: y = (2x + 3)(3x – 5) …(1)
Taking logarithm on both sides of (1), we get
log y = (3x – 5) log (2x + 3) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=(3 x-5) \cdot \frac{d}{d x}\{\log (2 x+3)\}+\log (2 x+3) \cdot \frac{d}{d x}(3 x-5)\)= \((3 x-5) \cdot \frac{1}{(2 x+3)} \cdot 2+\log (2 x+3) \cdot 3\)
⇒ \(\frac{d y}{d x}=y \cdot\left\{\frac{(6 x-10)}{(2 x+3)}+3 \log (2 x+3)\right\}\)
⇒ \(\frac{d y}{d x}=(2 x+3)^{(3 x-5)} \cdot\left\{\frac{(6 x-10)}{(2 x+3)}+3 \log (2 x+3)\right\} .\)
Example 8 If y = \(\frac{5^x}{x^5} \text {, find } \frac{d y}{d x} \text {. }\)
Solution
y = \(\frac{5^x}{x^5} \text {, find } \frac{d y}{d x} \text {. }\)
Given: y = \(\frac{5^x}{x^5}\) …(1)
Taking logarithm on both sides of (1), we get
log y = log(5x) – log(x5)
⇒ log y = x log 5 – 5 log x
⇒ \(\frac{1}{y} \cdot \frac{d y}{d x}=(\log 5) \cdot 1-\frac{5}{x}\) [differentiating both sides w.r.t. x]
⇒ \(\frac{d y}{d x}=y\left(\log 5-\frac{5}{x}\right)\)
⇒ \(\frac{d y}{d x}=\frac{5^x}{x^5}\left(\log 5-\frac{5}{x}\right) .\)
Example 9 Differentiate xx w.r.t. x.
Solution
xx w.r.t. x
Let y = xx …(1)
Taking logarithm on both sides of (1), we get
log y = x log x …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(x)\)= \(\left(x \cdot \frac{1}{x}+\log x \cdot 1\right)=(1+\log x)\)
⇒ \(\frac{d y}{d x}=y(1+\log x)\)
⇒ \(\frac{d y}{d x}=x^x(1+\log x) .\)
Example 10 Differentiate (sin x)x w.r.t. x.
Solution
(sin x)x w.r.t. x
Let y = (sin x)x …(1)
Taking logarithm on both sides of (1), we get
log y = x log(sin x) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{d}{d x}\{\log (\sin x)\}+\log (\sin x) \cdot \frac{d}{d x}(x)\)= \(x \cdot \frac{1}{\sin x} \cdot \cos x+\log (\sin x) \cdot 1\)
= x cot x + log (sin x)
⇒ \(\frac{d y}{d x}=y \cdot[x \cot x+\log (\sin x)]\)
⇒ \(\frac{d y}{d x}=(\sin x)^x[x \cot x+\log (\sin x)] .\)
Example 11 Differentiate xsin-1x w.r.t. x.
Solution
xsin-1x w.r.t. x
Let y = xsin-1x …(1)
Taking logarithm on both sides of (1), we get
log y = (sin-x)(log x) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\left(\sin ^{-1} x\right) \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)\)= \(\left(\sin ^{-1} x\right) \frac{1}{x}+(\log x) \cdot \frac{1}{\sqrt{1-x^2}}\)
⇒ \(\frac{d y}{d x}=y \cdot\left[\frac{\sin ^{-1} x}{x}+\frac{\log x}{\sqrt{1-x^2}}\right]\)
⇒ \(\frac{d y}{d x}=x^{\sin ^{-1} x} \cdot\left\{\frac{\sin ^{-1} x}{x}+\frac{\log x}{\sqrt{1-x^2}}\right\} .\)
Example 12 Differentiate (sin x)log x w.r.t. x.
Solution
(sin x)log x w.r.t. x
Let y = (sin x)log x …(1)
Taking logarithm on both sides of (1), we get
log y = (log x)(log sin x) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=(\log x) \cdot \frac{d}{d x}(\log \sin x)+(\log \sin x) \cdot \frac{d}{d x}(\log x)\)= \((\log x) \cdot \frac{1}{\sin x} \cdot \cos x+(\log \sin x) \cdot \frac{1}{x}\)
= \((\log x) \cot x+\frac{(\log \sin x)}{x}\)
⇒ \(\frac{d y}{d x}=y \cdot\left[(\log x) \cot x+\frac{(\log \sin x)}{x}\right]\)
⇒ \(\frac{d y}{d x}=(\sin x)^{\log x} \cdot\left[(\log x) \cot x+\frac{(\log \sin x)}{x}\right] .\)
Example 13 Differentiate xx sin-1√x w.r.t. x.
Solution
xx sin-1√x w.r.t. x
Let y = xx sin-1√x …(1)
Taking logarithm on both sides of (1), we get
log y = x log x + log(sin-1√x) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(x \log x)+\frac{d}{d x}\left\{\log \left(\sin ^{-1} \sqrt{x}\right)\right\}\)= \(\left(x \cdot \frac{1}{x}+\log x \cdot 1\right)+\frac{1}{\sin ^{-1} \sqrt{x}} \cdot \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2} x^{-1 / 2}\)
= \(\left\{(1+\log x)+\frac{1}{2\left(\sqrt{x-x^2}\right) \sin ^{-1} \sqrt{x}}\right\}\)
⇒ \(\frac{d y}{d x}=y \cdot\left\{(1+\log x)+\frac{1}{2\left(\sqrt{x-x^2}\right) \sin ^{-1} \sqrt{x}}\right\}\)
= \(\left(x^x \sin ^{-1} \sqrt{x}\right)\left\{(1+\log x)+\frac{1}{2\left(\sqrt{x-x^2}\right) \sin ^{-1} \sqrt{x}}\right\}\)
= \(\left\{\left(x^x \sin ^{-1} \sqrt{x}\right)(1+\log x)+\frac{x^x}{2\left(\sqrt{x-x^2}\right)}\right\} \text {. }\)
Example 14 Differentiate \(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\) w.r.t. x.
Solution
\(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\) w.r.t.
Let y = \(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\) …(1)
Taking logarithm on both sides of (1), we get
log y = x2 + log(tan-1x) – \(\frac{1}{2}\) log(1+x2) …(2)
On differentiating both sides of (2) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=2 x+\frac{1}{\tan ^{-1} x} \cdot \frac{1}{\left(1+x^2\right)}-\frac{1}{2} \cdot \frac{2 x}{\left(1+x^2\right)}\)⇒ \(\frac{d y}{d x}=y\left[2 x+\frac{1}{\left(1+x^2\right) \tan ^{-1} x}-\frac{x}{\left(1+x^2\right)}\right]\)
= \(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}},\left[2 x+\frac{1}{\left(1+x^2\right) \tan ^{-1} x}-\frac{x}{\left(1+x^2\right)}\right]\)
Example 15 Differentiate \(\left\{x^{\tan x}+\sqrt{\frac{x^2+1}{x}}\right\}\) w.r.t. x.
Solution
\(\left\{x^{\tan x}+\sqrt{\frac{x^2+1}{x}}\right\}\) w.r.t. x
Let y = v + u, where u = xtan x and v = \(\sqrt{\frac{x^2+1}{x}} .\)
Now, u = xtan x
⇒ log u = (tan x)(log x)
⇒ \(\frac{1}{u} \cdot \frac{d u}{d x}=(\tan x) \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(\tan x)\) [differentiating w.r.t. x]
= (tan x).\(\frac{1}{x}\) + (log x)sec2x
⇒ \(\frac{d u}{d x}=u \cdot\left[\frac{\tan x}{x}+(\log x) \sec ^2 x\right]\)
⇒ \(\frac{d u}{d x}=x^{\tan x} \cdot\left\{\frac{\tan x}{x}+(\log x) \sec ^2 x\right\}\) …(1)
And, v = \(\sqrt{\frac{x^2+1}{x}}\)
⇒ \(\frac{1}{v} \cdot \frac{d v}{d x}=\frac{1}{2} \cdot\left\{\frac{2 x}{\left(x^2+1\right)}-\frac{1}{x}\right\}\) [differentiating w.r.t. x]
⇒ \(\frac{d v}{d x}=\frac{v}{2} \cdot\left\{\frac{2 x^2-\left(x^2+1\right)}{x\left(x^2+1\right)}\right\}\)
⇒ \(\frac{d v}{d x}=\frac{1}{2} \sqrt{\frac{x^2+1}{x}} \cdot\left\{\frac{\left(x^2-1\right)}{x\left(x^2+1\right)}\right\}\) …(2)
∴ y = u + v
⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
⇒ \(\frac{d y}{d x}=x^{\tan x} \cdot\left\{\frac{\tan x}{x}+(\log x) \sec ^2 x\right\}+\frac{1}{2} \cdot \sqrt{\frac{x^2+1}{x}} \cdot\left\{\frac{\left(x^2-1\right)}{x\left(x^2+1\right)}\right\}\)
Example 16 If Y = (x)cos x + (cos x)sin x, find \(\frac{d x}{d y}\).
Solution
Y = (x)cos x + (cos x)sin x
Let y = u + v, where u = (x)cos x and v = (cos x)sin x.
Now, u = (x)cos x
⇒ log u = (cos x)(log x)
⇒ \(\frac{1}{u} \cdot \frac{d u}{d x}=(\cos x) \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(\cos x)\) [on differentiating w.r.t. x]
= (cos x).\(\frac{1}{x}\) + (log x)(-sin x)
⇒ \(\frac{d u}{d x}=u \cdot\left\{\frac{\cos x}{x}-(\log x)(\sin x)\right\}\)
⇒ \(\frac{d u}{d x}=(x)^{\cos x}\left\{\frac{\cos x}{x}-(\log x)(\sin x)\right\}\) …(1)
And, v = (cos x)sin x
⇒ log v = (sin x)log(cos x)
⇒ \(\frac{1}{v} \cdot \frac{d v}{d x}=(\sin x) \cdot \frac{d}{d x}\{\log (\cos x)\}+\log (\cos x) \cdot \frac{d}{d x}(\sin x)\) [on differentiating w.r.t. x]
⇒ \(\frac{d v}{d x}=v \cdot\left\{(\sin x) \cdot \frac{(-\sin x)}{\cos x}+\log (\cos x) \cdot \cos x\right\}\)
⇒ \(\frac{d v}{d x}=(\cos x)^{\sin x} \cdot\{-\sin x \tan x+\cos x \cdot \log (\cos x)\}\) …(2)
∴ y = (u+v)
⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
⇒ \(\frac{d y}{d x}=(x)^{\cos x} \cdot\left\{\frac{\cos x}{x}-(\log x) \sin x\right\}+(\cos x)^{\sin x} \cdot\{-\sin x \tan x+\cos x \cdot \log (\cos x)\}\)
Example 17 If y = (sin x)tan x + (cos x)sec x, find \(\frac{d x}{d y}\).
Solution
y = (sin x)tan x + (cos x)sec x
Let y = u + v, where u = (sin x)tan x and v = (cos x)sec x.
Now, u = (sin x)tan x
⇒ log u = (tan x)(log sin x)
⇒ \(\frac{1}{u} \cdot \frac{d u}{d x}=(\tan x) \cdot \frac{d}{d x}(\log \sin x)+(\log \sin x) \cdot \frac{d}{d x}(\tan x)\) [on differentiating w.r.t. x]
= \((\tan x) \cdot \frac{1}{\sin x} \cdot \cos x+(\log \sin x) \cdot \sec ^2 x\)
⇒ \(\frac{d u}{d x}=u\left[1+(\log \sin x) \sec ^2 x\right]\)
⇒ \(\frac{d u}{d x}=(\sin x)^{\tan x} \cdot\left\{1+(\log \sin x) \sec ^2 x\right\}\) …(1)
And v = (cos x)sec x
⇒ log v = (sec x).log(cos x)
⇒ \(\frac{1}{v} \cdot \frac{d v}{d x}=(\sec x) \cdot \frac{d}{d x}\{\log (\cos x)\}+\log (\cos x) \cdot \frac{d}{d x}(\sec x)\) [on differentiating w.r.t. x]
= \((\sec x) \cdot \frac{1}{\cos x} \cdot(-\sin x)+\log (\cos x) \cdot \sec x \tan x\)
= (sec x tan x)[log (cos x) – 1]
⇒ \(\frac{d v}{d x}=v \cdot(\sec x \tan x)[\log (\cos x)-1]\)
⇒ \(\frac{d v}{d x}=(\cos x)^{\sec x} \cdot(\sec x \tan x)[\log (\cos x)-1]\) …(2)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
⇒ \(\frac{d y}{d x}=(\sin x)^{\tan x} \cdot\left\{1+(\log \sin x) \sec ^2 x\right\}+(\cos x)^{\sec x} \cdot(\sec x \tan x)[\log (\cos x)-1]\) [from (1) and (2)].
Example 18 If xy = yx, find \(\frac{d x}{d y}\).
Solution
xy = yx
Given: xy = yx
⇒ y log x = x log y …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(y \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(y)=x \cdot \frac{d}{d x}(\log y)+(\log y) \cdot \frac{d}{d x}(x)\)⇒ \(y \cdot \frac{1}{x}+(\log x) \cdot \frac{d y}{d x}=x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+(\log y) \cdot 1\)
⇒ \(\left(\log x-\frac{x}{y}\right) \frac{d y}{d x}=\left(\log y-\frac{y}{x}\right)\)
⇒ \(\frac{(y \log x-x)}{y} \cdot \frac{d y}{d x}=\frac{(x \log y-y)}{x}\)
⇒ \(\frac{d y}{d x}=\frac{y(x \log y-y)}{x(y \log x-x)}\)
Example 19 If xy . yx = 1, find \(\frac{d x}{d y}\).
Solution
xy . yx = 1
Given: xy . yx = 1
⇒ (y log x) + (x log y) = 0 …(1) [∵ log 1 = 0]
On differentiating both sides of (1) w.r.t. x, we get
\(y \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(y)+x \cdot \frac{d}{d x}(\log y)+(\log y) \cdot \frac{d}{d x}(x)=0\)⇒ \(y \cdot \frac{1}{x}+(\log x) \cdot \frac{d y}{d x}+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+(\log y) \cdot 1=0\)
⇒ \(\left(\log x+\frac{x}{y}\right) \cdot \frac{d y}{d x}=-\left(\log y+\frac{y}{x}\right)\)
⇒ \(\frac{(y \log x+x)}{y} \cdot \frac{d y}{d x}=\frac{-(x \log y+y)}{x}\)
⇒ \(\frac{d y}{d x}=\frac{-y(x \log y+y)}{x(y \log x+x)} .\)
Example 20 If xy = ex-y, prove that \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^2}\).
Solution
xy = ex-y
We have
xy = ex-y ⇒ y log x = (x-y)
⇒ (1 + log x)y = x
⇒ \(y=\frac{x}{(1+\log x)}\) …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\frac{d y}{d x}=\frac{(1+\log x) \cdot \frac{d}{d x}(x)-x \cdot \frac{d}{d x}(1+\log x)}{(1+\log x)^2}\)= \(\frac{(1+\log x) \cdot 1-x \cdot \frac{1}{x}}{(1+\log x)^2}=\frac{(1+\log x-1)}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}\)
Example 21 If xy + yx = ab, find \(\frac{d x}{d y}\).
Solution
xy + yx = ab
Let u = xy and v = yx.
Then, u + v = ab ⇒ \(\frac{d u}{d x}+\frac{d v}{d x}=0\) …(1) [∵ ab = constant]
Now, u = xy ⇒ log u = y log x [taking log on both sides]
⇒ \(\frac{1}{u} \cdot \frac{d u}{d x}=y \cdot \frac{1}{x}+\log x \cdot \frac{d y}{d x}\) [on differentiation]
⇒ \(\frac{d u}{d x}=u\left[\frac{y}{x}+\log x \cdot \frac{d y}{d x}\right]\)
⇒ \(\frac{d u}{d x}=x^y\left[\frac{y+x \log x \cdot \frac{d y}{d x}}{x}\right]\)
⇒ \(\frac{d u}{d x}=x^{y-1}\left[y+x \log x \cdot \frac{d y}{d x}\right]\)
And, v = yx ⇒ log v = x log y [taking log on both sides]
⇒ \(\frac{1}{v} \cdot \frac{d v}{d x}=x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y\) [on differentiation]
⇒ \(\frac{d v}{d x}=v \cdot\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right] \Rightarrow \frac{d v}{d x}=y^x\left\{\frac{x \cdot \frac{d y}{d x}+y \log y}{y}\right\}\)
⇒ \(\frac{d v}{d x}=y^{(x-1)},\left\{x \cdot \frac{d y}{d x}+y \log y\right\} .\)
Using (1), we get \(\frac{d u}{d x}+\frac{d v}{d x}=0\)
⇒ \(x^{(y-1)}\left\{y+x \log x \cdot \frac{d y}{d x}\right\}+y^{(x-1)} \cdot\left\{x \frac{d y}{d x}+y \log y\right\}=0\)
⇒ \(\left\{x^y(\log x)+x \cdot y^{(x-1)}\right\} \cdot \frac{d y}{d x}=-\left\{y \cdot x^{(y-1)}+y^x(\log y)\right\}\)
∴ \(\frac{d y}{d x}=\frac{-\left\{y \cdot x^{(y-1)}+y^x(\log y)\right\}}{\left\{x^y(\log x)+x y^{(x-1)}\right\}} .\)
Example 22 If xx + xy + yx = ab, find \(\frac{d x}{d y}\).
Solution
xx + xy + yx = ab
Let u = xx, v = xy and w = yx. Then,
u + v + w = ab
⇒ \(\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}=0\) …(1) [∵ ab = constant]
Now, u = xx
⇒ log u = x log x
⇒ \(\frac{1}{u} \cdot \frac{d u}{d x}=x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(x)\) [on differentiating both sides w.r.t. x]
⇒ \(\frac{d u}{d x}=u\left[\frac{y}{x}+\log x \cdot \frac{d y}{d x}\right]\)
⇒ \(\frac{d u}{d x}=x^y\left[\frac{y+x \log x \cdot \frac{d y}{d x}}{x}\right]\) …(2)
And, v = xy
⇒ log v = y log x
⇒ \(\frac{1}{v}, \frac{d v}{d x}=y \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(y)\) [on differentiating both sides w.r.t. x]
⇒ \(\frac{d v}{d x}=v \cdot\left\{y \cdot \frac{1}{x}+(\log x) \cdot \frac{d y}{d x}\right\}\)
⇒ \(\frac{d v}{d x}=x^y\left\{\frac{y}{x}+(\log x) \frac{d y}{d x}\right\}\) …(3)
And, w = yx
⇒ log w = x log y
⇒ \(\frac{1}{w} \cdot \frac{d w}{d x}=x \cdot \frac{d}{d x}(\log y)+(\log y) \cdot \frac{d}{d x}(x)\) [on differentiating both sides w.r.t. x]
⇒ \(\frac{d w}{d x}=w \cdot\left\{x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+(\log y) \cdot 1\right\}\)
⇒ \(\frac{d w}{d x}=y^x \cdot\left\{\frac{x}{y} \cdot \frac{d y}{d x}+(\log y)\right\}\) …(4)
Using (2), (3) and (4) in (1), we get
\(x^x(1+\log x)+x^y\left\{\frac{y}{x}+(\log x) \frac{d y}{d x}\right\}+y^x \cdot\left\{\frac{x}{y} \cdot \frac{d y}{d x}+(\log y)\right\}=0\)⇒ \(\left\{x^x(1+\log x)+y \cdot x^{(y-1)}+y^x(\log y)\right\}+\left\{x^y(\log x)+x y^{(x-1)}\right\} \frac{d y}{d x}=0\)
⇒ \(\frac{d y}{d x}=\frac{-\left\{x^x(1+\log x)+y \cdot x^{(y-1)}+y^z(\log y)\right\}}{\left\{x^y(\log x)+x y^{(x-1)}\right\}} .\)
Example 23 If y = \(x^{\left(x^x\right)}\), find [/latex]\frac{d x}{d y}[/latex].
Solution
y = \(x^{\left(x^x\right)}\)
Let xx = u . Then, y = xu.
∴ x log x = log u and log y = u log x.
Now, log u = x log x
⇒ \(\frac{1}{u} \cdot \frac{d u}{d x}=x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(x)\) [on differentiating w.r.t. x]
⇒ \(\frac{d u}{d x}=u \cdot\left[x \cdot \frac{1}{x}+\log x \cdot 1\right]\)
⇒ \(\frac{d u}{d x}=x^x(1+\log x)\) …(1)
And, log y = u log x
⇒ \(\frac{1}{y} \cdot \frac{d y}{d x}=u \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(u)\)
= \(u \cdot \frac{1}{x}+(\log x) \cdot \frac{d u}{d x}\)
⇒ \(\frac{d y}{d x}=y \cdot\left[\frac{u}{x}+(\log x) \cdot \frac{d u}{d x}\right]\)
= \(x^{\left(x^x\right)} \cdot\left[\frac{x^x}{x}+(\log x)\left\{x^x(1+\log x)\right\}\right]\)
= \(x^{\left(x^x\right)} \cdot\left[x^{(x-1)}+x^x(\log x)+x^x(\log x)^2\right]\).
Example 24 If y = \((\log x)^{\cos x}+\frac{x^2+1}{x^2-1} \text {, find } \frac{d y}{d x}\)
Solution
y = \((\log x)^{\cos x}+\frac{x^2+1}{x^2-1} \text {, find } \frac{d y}{d x}\)
Let (log x)cos x = u and and \(\frac{x^2+1}{x^2-1}\) = v. Then,
u = (log x)cos x ⇒ log u = cos x . log(log x) …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\frac{1}{u} \cdot \frac{d u}{d x}=\cos x \cdot \frac{d}{d x}\{\log (\log x)\}+\log (\log x) \cdot \frac{d}{d x}(\cos x)\)= \(\cos x \frac{1}{\log x} \cdot \frac{1}{x}-(\sin x) \cdot \log (\log x)\)
∴ \(\frac{d u}{d x}=u \cdot\left\{\frac{\cos x}{x \log x}-(\sin x) \cdot \log (\log x)\right\}\)
⇒ \(\frac{d u}{d x}=(\log x)^{\cos x} \cdot\left\{\frac{\cos x}{x \log x}-(\sin x) \cdot \log (\log x)\right\} \text {. }\)
Also, v = \(\frac{\left(x^2+1\right)}{\left(x^2-1\right)} ⇒ \frac{d v}{d x}=\frac{\left(x^2-1\right) \cdot \frac{d}{d x}\left(x^2+1\right)-\left(x^2+1\right) \cdot \frac{d}{d x}\left(x^2-1\right)}{\left(x^2-1\right)^2}\)
⇒ \(\frac{d v}{d x}=\frac{\left(x^2-1\right) \cdot 2 x-\left(x^2+1\right) \cdot 2 x}{\left(x^2-1\right)^2}=\frac{-4 x}{\left(x^2-1\right)^2}\)
∴ y = u + v
⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
= \((\log x)^{\cos x} \cdot\left\{\frac{\cos x}{x \log x}-(\sin x) \cdot \log (\log x)\right\}-\frac{4 x}{\left(x^2-1\right)^2}\)
7. Derivatives of an Infinite Series
If we take out a single term from an infinite series, it remains unaffected. We utilize this result in finding the derivative of an infinite series.
Solved Examples
Example 1 If \(y=x^{x^{x \cdots \cdots}}, \text { prove that } \frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\).
Solution
\(y=x^{x^{x \cdots \cdots}}, \text { prove that } \frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\).
We know that an infinite series is not affected by the exclusion of a single term.
So, we may write the given function as y = xy.
Now, y = xy ⇒ log y = y log x …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{1}{x}+\log x \cdot \frac{d y}{d x}\)⇒ \(\left(\frac{1}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\frac{(1-y \log x)}{y} \cdot \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\frac{d y}{d x}=\left\{\frac{y}{x} \times \frac{y}{(1-y \log x)}\right\} \Rightarrow \frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\)
Example 2 If \(y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\ldots \text { to } \infty}}} \text {, prove that } \frac{d y}{d x}=\frac{\cos x}{(2 y-1)} \text {. }\)
Solution
\(y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\ldots \text { to } \infty}}} \text {, prove that } \frac{d y}{d x}=\frac{\cos x}{(2 y-1)} \text {. }\)We may write the given series as
\(y=\sqrt{\sin x+y} \Rightarrow y^2=(\sin x+y)\) …(1)
On differentiating both sides of (1), w.r.t. x, we get
\(2 y \cdot \frac{d y}{d x}=\cos x+\frac{d y}{d x}\)⇒ \((2 y-1) \cdot \frac{d y}{d x}=\cos x\)
⇒ \(\frac{d y}{d x}=\frac{\cos x}{(2 y-1)} .\)
Example 3 If \(y=e^{x+e^{x+e^{x+\ldots infinity}}}, \text { prove that } \frac{d y}{d x}=\frac{y}{(1-y)} .\)
Solution
\(y=e^{x+e^{x+e^{x+\ldots infinity}}}, \text { prove that } \frac{d y}{d x}=\frac{y}{(1-y)} .\)We may write the given series as
y = ex+y ⇒ log y = (x + y) …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=1+\frac{d y}{d x}\)⇒ \(\left(\frac{1}{y}-1\right) \frac{d y}{d x}=1\)
⇒ \(\frac{(1-y)}{y}, \frac{d y}{d x}=1\)
⇒ \(\frac{d y}{d x}=\frac{y}{(1-y)} .\)
Example 4 If \(y=(\sqrt{x})^{(\sqrt{x})^{(\sqrt{x}) \ldots infinity}}\), prove that \(x\left(\frac{d y}{d x}\right)=\frac{y^2}{(2-y \log x)}\).
Solution
\(y=(\sqrt{x})^{(\sqrt{x})^{(\sqrt{x}) \ldots infinity}}\), prove that \(x\left(\frac{d y}{d x}\right)=\frac{y^2}{(2-y \log x)}\).
We may write the given series as
\(y=(\sqrt{x})^y \Rightarrow y=x^{y / 2}\)⇒ \(\log y=\frac{y}{2} \cdot \log x\) …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=\frac{y}{2} \cdot \frac{1}{x}+\frac{1}{2} \log x \cdot \frac{d y}{d x}\)⇒ \(\left(\frac{1}{y}-\frac{1}{2} \log x\right) \cdot \frac{d y}{d x}=\frac{y}{2 x}\)
⇒ \(\frac{(2-y \log x)}{2 y} \cdot \frac{d y}{d x}=\frac{y}{2 x}\)
⇒ \(x \cdot \frac{d y}{d x}=\frac{y^2}{(2-y \log x)}\)
8. Derivatives of One Function With Respect To Another Function
Let f(x) and g(x) be two functions of x. In order to find the derivative of f(x) with respect to g(x), we put u = f(x) and v = g(x). Now, find \frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}, which is the required derivative.
Example 1 Differentiate ex w.r.t. √x.
Solution
ex w.r.t. √x
Let u = ex and v = √x.
Then, \(\frac{d u}{d x}=e^x \text { and } \frac{d v}{d x}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}} .\)
∴ \(\frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{e^x}{(1 / 2 \sqrt{x})}=2 e^x \sqrt{x}\)
Example 2 Differentiate sin2x w.r.t. ecos x.
Solution
sin2x w.r.t. ecos x
Let u = sin2x and v = ecos x. Then,
\(\frac{d u}{d x}=2 \sin x \cos x\)and \(\frac{d v}{d x}=e^{\cos x} \cdot(-\sin x)=(-\sin x) \cdot e^{\cos x}\)
∴ \(\frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{2 \sin x \cos x}{(-\sin x) e^{\cos x}}=\frac{-2 \cos x}{e^{\cos x}} .\)
Example 3 Differentiate sin-1x w.r.t. tan-1x.
Solution
sin-1x w.r.t. tan-1x
Let u = sin-1x and v = tan-1x.
Then, \(\frac{d u}{d x}=\frac{1}{\sqrt{1-x^2}} \text { and } \frac{d v}{d x}=\frac{1}{\left(1+x^2\right)}\)
∴ \(\frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{1}{\sqrt{1-x^2}} \times\left(1+x^2\right)=\frac{\left(1+x^2\right)}{\sqrt{1-x^2}}\)
Example 4 Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) w.r.t. tan-1x.
Solution
\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) w.r.t. tan-1x
Let u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) and v = tan-1x.
Now, v = tan-1x ⇒ x = tan v.
Putting x = tan v, we get
u = \(\tan ^{-1}\left\{\frac{\sqrt{1+\tan ^2 v}-1}{\tan v}\right\}=\tan ^{-1}\left(\frac{\sec v-1}{\tan v}\right)\)
= \(\tan ^{-1}\left(\frac{1-\cos v}{\sin v}\right)=\tan ^{-1}\left\{\frac{2 \sin ^2(\nu / 2)}{2 \sin (v / 2) \cos (v / 2)}\right\}\)
= \(\tan ^{-1}\left\{\tan \frac{v}{2}\right\}=\frac{v}{2}\)
∴ \(u=\frac{v}{2} \Rightarrow \frac{d u}{d v}=\frac{1}{2}\)
Example 5 Differentiate \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \text { w.r.t. } \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
Solution
\(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \text { w.r.t. } \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)Let u = \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) and v = \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
Putting x = tan θ, we get
u = \(\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta \text {, }\)
v = \(\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)=\cos ^{-1}(\cos 2 \theta)=2 \theta \text {. }\)
∴ u = v ⇒ \(\frac{d u}{d v}=1\).
Example 6 Differentiate \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \cdot \text { w.r.t. } \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \text {. }\)
Solution
\(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \cdot \text { w.r.t. } \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \text {. }\)Let u = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) and v = \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Putting x = tan θ, we get
u = \(\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)=\tan ^{-1}(\tan 2 \theta)=2 \theta\),
v = \(\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta \text {. }\)
∴ u = v ⇒ \(\frac{d u}{d v}=1\).
Example 7 Differentiate \(\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right\} \text { w.r.t. } \cos ^{-1} x^2\)
Solution
\(\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right\} \text { w.r.t. } \cos ^{-1} x^2\)Let u = \(\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right\}\) and v = cos-1x2.
Then, cos-1x2 = v ⇒ x2 = cos v.
Putting x2 = cos v, we get
u = \(\tan ^{-1}\left\{\frac{\sqrt{1+\cos v}-\sqrt{1-\cos v}}{\sqrt{1+\cos v}+\sqrt{1-\cos v}}\right\}\)
= \(\tan ^{-1}\left\{\frac{\sqrt{2 \cos ^2(v / 2)}-\sqrt{2 \sin ^2(v / 2)}}{\sqrt{2 \cos ^2(v / 2)}+\sqrt{2 \sin ^2(v / 2)}}\right\}\)
= \(\tan ^{-1}\left\{\frac{\cos (v / 2)-\sin (v / 2)}{\cos (v / 2)+\sin (v / 2)}\right\}=\tan ^{-1}\left\{\frac{1-\tan (v / 2)}{1+\tan (v / 2)}\right\}\)
[dividing num. and denom. by cos(v/2)]
= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{v}{2}\right)\right\}=\left(\frac{\pi}{4}-\frac{v}{2}\right) .\)
∴ \(u=\left(\frac{\pi}{4}-\frac{v}{2}\right) \Rightarrow \frac{d u}{d \nu}=\frac{-1}{2}\)
10. Second-Order Derivatives
Let f(x) be a differentiable function of x whose second-order derivative exists. We denote the second-order derivative of y w.r.t. x by \frac{d^2 y}{d x^2} or y2.
Solved Examples
Example 1 Find the second-order derivative of:
(1) x10
(2) log x
(3) tan-1x
Solution
(1) Let y = x10. Then,
\(\frac{d y}{d x}=10 x^9 .\)∴ \(\frac{d^2 y}{d x^2}=(10 \times 9) x^8=90 x^8 .\)
Hence, \(\frac{d^2}{d x^2}\left(x^{10}\right)=90 x^8\)
(2) Let y = log x. Then,
\(\frac{d y}{d x}=\frac{1}{x}=x^{-1} \text {. }\)∴ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(x^{-1}\right)=(-1) x^{(-1-1)}=-x^{-2}=\frac{-1}{x^2} .\)
Hence, \(\frac{d^2}{d x^2}(\log x)=\frac{-1}{x^2}\)
(3) Let y = tan-1x. Then,
\(\frac{d y}{d x}=\frac{1}{\left(1+x^2\right)}=\left(1+x^2\right)^{-1} .\)∴ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(1+x^2\right)^{-1}\)
= \((-1)\left(1+x^2\right)^{-2} \cdot(2 x)=\frac{-2 x}{\left(1+x^2\right)^2}\)
Hence, \(\frac{d^2}{d x^2}\left\{\tan ^{-1} x\right\}=\frac{-2 x}{\left(1+x^2\right)^2}\).
Example 2 If y = (tan x + sec x), prove that \(\frac{d^2 y}{d x^2}=\frac{\cos x}{(1-\sin x)^2}\)
Solution
Given that y = (tan x + sec x).
∴ \(\frac{d y}{d x}=\sec ^2 x+\sec x \tan x=\left(\frac{1}{\cos ^2 x}+\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}\right)\)
= \(\left(\frac{1+\sin x}{\cos ^2 x}\right)=\left(\frac{1+\sin x}{1-\sin ^2 x}\right)=\frac{1}{(1-\sin x)}\)
∴ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left\{\frac{1}{(1-\sin x)}\right\}=\frac{d}{d x}(1-\sin x)^{-1}\)
= \((-1)(1-\sin x)^{-2}(-\cos x)=\frac{\cos x}{(1-\sin x)^2} \text {. }\)
Hence, \(\frac{d^2 y}{d x^2}=\frac{\cos x}{(1-\sin x)^2}\)
Example 3 If y = e4xsin 3x, find \(\frac{d^2 y}{d x^2} .\)
Solution
Let y = e4x sin 3x. Then,
\(\frac{d y}{d x}=3 e^{4 x} \cos 3 x+4 e^{4 x} \sin 3 x=e^{4 x}(3 \cos 3 x+4 \sin 3 x) .\)∴ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left\{e^{4 x}(3 \cos 3 x+4 \sin 3 x)\right\}\)
= e4x (-9sin 3x + 12 cos 3x) + 4 e4x (3 cos 3x + 4 sin 3x)
= e4x (7 sin 3x + 24 cos 3x).
\(\frac{d^2 y}{d x^2} .\) = e4x (7 sin 3x + 24 cos 3x).
Example 4 If y = (x4 + cot x), find \(\frac{d^2 y}{d x^2}\).
Solution
We have
y = (x4 + cot x)
⇒ \(\frac{d y}{d x}=4 x^3-{cosec}^2 x\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(4 x^3-{cosec}^2 x\right)\)
= \(4 \cdot \frac{d}{d x}\left(x^3\right)-\frac{d}{d x}\left({cosec}^2 x\right)\)
= (4 x 3x2) – 2 cosec x(-cosec x cot x)
= (12 x2 + 2 cosec2x cot x).
\(\frac{d^2 y}{d x^2}\) = (12 x2 + 2 cosec2x cot x).
Example 5 Find the second derivative of log(log x) w.r.t. x.
Solution
Let y = log(log x). Then,
\(\frac{d y}{d x}=\frac{1}{(\log x)} \cdot \frac{1}{x}=\frac{1}{(x \log x)}=(x \log x)^{-1}\)⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}(x \log x)^{-1}\)
= \((-1)(x \log x)^{-2} \cdot \frac{d}{d x}(x \log x)\)
= \(\frac{-1}{(x \log x)^2} \cdot\left(x \cdot \frac{1}{x}+\log x \cdot 1\right)=\frac{-(1+\log x)}{(x \log x)^2} .\)
log(log x) = \(\frac{-1}{(x \log x)^2} \cdot\left(x \cdot \frac{1}{x}+\log x \cdot 1\right)=\frac{-(1+\log x)}{(x \log x)^2} .\)
Example 6 If y = sin(log x), find \(\frac{d^2 y}{d x^2}\).
Solution
We have
y = sin(log x)
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\{\sin (\log x)\}=\cos (\log x) \cdot \frac{1}{x}=\frac{\cos (\log x)}{x}\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x} \cdot\left\{\frac{\cos (\log x)}{x}\right\}\)
= \(\frac{x \cdot \frac{d}{d x}\{\cos (\log x)\}-\cos (\log x) \cdot \frac{d}{d x}(x)}{x^2}\)
= \(\frac{x\left\{-\sin (\log x) \cdot \frac{1}{x}\right\}-\cos (\log x) \cdot 1}{x^2}\)
= \(\frac{-\{\sin (\log x)+\cos (\log x)\}}{x^2} .\)
\(\frac{d^2 y}{d x^2}\) = \(\frac{-\{\sin (\log x)+\cos (\log x)\}}{x^2} .\)
Example 7 If ey(x+1) = 1, prove that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\)
Solution
We have
ey(x+1) = 1 ⇒ \(e^y=\frac{1}{(x+1)}\) …(1)
⇒ \(y=\log \left\{\frac{1}{(x+1)}\right\}=\log 1-\log (x+1)\)
⇒ y = -log(x+1) …(2)
∴ \(\frac{d y}{d x}=\frac{-1}{(x+1)}\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{1}{(x+1)^2}=\left(\frac{d y}{d x}\right)^2 \text {. }\)
Hence, \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\)
Example 8 If y = A cos nx + B sin nx, prove that \(\frac{d^2 y}{d x^2}+n^2 y=0\)
Solution
We have
y = A cos nx + B sin nx
⇒ \(\frac{d y}{d x}=\frac{d}{d x}(A \cos n x)+\frac{d}{d x}(B \sin n x)\)
= -An sin nx + B n cos nx
= n(B cos nx – A sin nx)
⇒ \(\frac{d^2 y}{d x^2}=n \cdot \frac{d}{d x}(B \cos n x-A \sin n x)\)
= \(n \cdot\left\{B \cdot \frac{d}{d x}(\cos n x)-A \cdot \frac{d}{d x}(\sin n x)\right\}\)
= n . {-B n sin nx – A n cos nx}
= -n2(A cos nx + B sin nx) = -n2y
⇒ \(\frac{d^2 y}{d x^2}+n^2 y=0\)
Example 9 If y = ex(sin x + cos x), prove that \(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0\)
Solution
We have
y = ex(sin x + cos x)
⇒ \(\frac{d y}{d x}=e^x \cdot \frac{d}{d x}(\sin x+\cos x)+(\sin x+\cos x) \cdot \frac{d}{d x}\left(e^x\right)\)
= ex(cos x – sin x) + (sin x + cos x) . ex = 2 ex cos x
⇒ \(\frac{d^2 y}{d x^2}=2 \cdot \frac{d}{d x}\left(e^x \cos x\right)\)
= \(2 \cdot\left\{e^x \cdot \frac{d}{d x}(\cos x)+\cos x \cdot \frac{d}{d x}\left(e^x\right)\right\}\)
= 2 . {ex(-sin x) + (cos x)ex} = 2ex(cos x – sin x).
∴ \(\left(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y\right)\)
= 2 ex (cos x – sin x) – 4 ex cos x + 2 ex(sin x + cos x) = 0.
Hence, \(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0\)
Example 10 If y = 3 e2x + 2 e3x, prove that \(\frac{d^2 y}{d x^2}-5 \frac{d y}{d x}+6 y=0 .\)
Solution
We have
y = 3 e2x + 2 e3x …(1)
⇒ \(\frac{d y}{d x}=3 \cdot \frac{d}{d x}\left(e^{2 x}\right)+2 \cdot \frac{d}{d x}\left(e^{3 x}\right)\) [on differentiating (1) w.r.t. x]
= (3 x 2e2x) + (2 x 3 e3x) = (6 e2x + 6 e3x)
⇒ \(\frac{d y}{d x}=6\left(e^{2 x}+e^{3 x}\right)\) …(2)
On differentiating (2) w.r.t. x, we get
\(\frac{d^2 y}{d x^2}=6 \cdot\left\{\frac{d}{d x}\left(e^{2 x}\right)+\frac{d}{d x}\left(e^{3 x}\right)\right\}\)= 6 . (2 e2x + e3x)
∴ \(\left(\frac{d^2 y}{d x^2}-5 \frac{d y}{d x}+6 y\right)\)
= 6(2 e2x + 3 e3x) – 30(e2x + e3x) + (18 e2x + 12 e3x)
= (12 – 30 + 18)e2x + (18 – 30 + 12)e3x = 0.
Hence, \(\left(\frac{d^2 y}{d x^2}-5 \frac{d y}{d x}+6 y\right)=0 .\)
Example 11 If y = sin-1x, prove that \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=0 .\)
Solution
Given: y = sin-1x …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(y_1=\frac{1}{\sqrt{1-x^2}}\)⇒ \(y_1^2=\frac{1}{\left(1-x^2\right)}\)
⇒ (1 – x2)y12 = 1 …(2)
On differentiating both sides of (2) w.r.t. x, we get
(1-x2) . 2y1y2 + y121(-2x) = 0
⇒ (1-x2)y2 – xy1 = 0.
Hence, \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=0 .\)
Example 12 If y = (tan-1x)2, prove that (1+x2)2y2 + 2x(1+x2)y1 = 2.
Solution
Given: y = (tan-1x)2 …(1)
On differentiating both sides of (1) w.r.t. x, we get
\(y_1=2 \tan ^{-1} x \cdot \frac{1}{\left(1+x^2\right)}\)⇒ (1 + x2)y1 = 2 tan-1x
⇒ (1 + x2)2y12 = 4(tan-1x)2 [on squaring both sides]
⇒ (1 + x2)2y12 – 4y = 0 …(2)
On differentiating both sides of (2) w.r.t. x, we get
(1+x2)2. 2y1y2 + y12 . 2(1+x2).2x – 4y1 = 0
⇒ (1+x2)2y2 + 2x(1+x2)y1 – 2 = 0.
Hence, (1+x2)2y2 + 2x(1+x2)y1 = 2.
Example 13 If x = a(θ + sin θ) and y = a(1 – cos θ), find \(\frac{d^2 y}{d x^2} \text { at } \theta=\frac{\pi}{2}\).
Solution
We have
x = a(θ + sin θ) and y = a(1 – cos θ)
⇒ \(\frac{d x}{d \theta}=a(1+\cos \theta) \text { and } \frac{d y}{d \theta}=a \sin \theta\)
⇒ \(\frac{d y}{d x}=\frac{(d y / d \theta)}{(d x / d \theta)}\)
= \(\frac{a \sin \theta}{a(1+\cos \theta)}=\frac{\sin \theta}{(1+\cos \theta)}=\frac{2 \sin (\theta / 2) \cos (\theta / 2)}{2 \cos ^2(\theta / 2)}=\tan \frac{\theta}{2}\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\tan \frac{\theta}{2}\right)=\frac{1}{2} \sec ^2 \frac{\theta}{2} \cdot \frac{d \theta}{d x}=\left(\frac{1}{2} \sec ^2 \frac{\theta}{2}\right) \times \frac{1}{a(1+\cos \theta)}\)
⇒ \(\left(\frac{d^2 y}{d x^2}\right)_{\theta=\frac{\pi}{2}}=\frac{1}{2} \sec ^2 \frac{\pi}{4} \cdot \frac{1}{a\left(1+\cos \frac{\pi}{2}\right)}=\frac{1}{a} .\)
Example 14 If x = (2 cos θ – cos 2θ) and y = (2 sin θ – sin 2θ), find \(\frac{d^2 y}{d x^2} \text { at } \theta=\frac{\pi}{2}\).
Solution
We have
x = (2 cos θ – cos 2θ) and y = (2 sin θ – sin 2θ)
⇒ \(\frac{d x}{d \theta}=(-2 \sin \theta+2 \sin 2 \theta) \text { and } \frac{d y}{d \theta}=(2 \cos \theta-2 \cos 2 \theta)\)
⇒ \(\frac{d y}{d x}=\frac{(d y / d \theta)}{(d x / d \theta)}=\frac{(2 \cos \theta-2 \cos 2 \theta)}{(-2 \sin \theta+2 \sin 2 \theta)}=\frac{(\cos \theta-\cos 2 \theta)}{(\sin 2 \theta-\sin \theta)}\)
= \(\frac{2 \sin \left(\frac{3 \theta}{2}\right) \sin \frac{\theta}{2}}{2 \cos \left(\frac{3 \theta}{2}\right) \sin \frac{\theta}{2}}=\tan \frac{3 \theta}{2}\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\tan \frac{3 \theta}{2}\right)=\frac{3}{2} \sec ^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}=\frac{3}{2} \sec ^2 \frac{3 \theta}{2} \cdot \frac{1}{2(\sin 2 \theta-\sin \theta)}\)
⇒ \(\left(\frac{d^2 y}{d x^2}\right)_{\theta=\frac{\pi}{2}}=\frac{3}{2} \cdot \sec ^2\left(\frac{3 \pi}{4}\right) \cdot \frac{1}{2\left(\sin \pi-\sin \frac{\pi}{2}\right)}\)
= \(\frac{-3}{4} \sec ^2 \frac{\pi}{4}=\frac{-3}{4} \times(\sqrt{2})^2=\frac{-3}{2}\)
[∵ \(\sec \frac{3 \pi}{4}=\sec \left(\pi-\frac{\pi}{4}\right)=-\sec \frac{\pi}{4}\)].