# First Order Derivative Solution – Part I

Differential Calculus / Sunday, August 19th, 2018

Table of Contents

# Introduction:

The main or central concept of Differential Calculus is the idea of derivative or differential coefficient of a function. While formulating the laws of motion in Mechanics, Sir Isaac Newton introduced the idea of rate of change of momentum and he termed this rate of change as flux. In his treatise ‘Theory of Fluxion’, Newton made many significant discussions on the measurement of the rate of change of a variable quantity. This famous book laid the foundation of Differential Calculus.

## Derivative or Differential Coefficient of a Function

Let y = f(x) be function defined over an interval containing the point x=a. Let x be given an increment h or ∆x at this point (h may be positive or negative); and let ∆y be the corresponding increment of y.
If the limit,
$\underset{h\to 0}{\mathop{\lim }}\,\frac{\Delta y}{\Delta x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(a+h)-f(a)}{h}$
exists, f(x) is said to be derivable at x=a. the value of this limit is called the derivative of f(x) at x=a. This is denoted by,
$f'(a)or\left[ \frac{d}{dx}\left\{ f(x) \right\} \right]or{{\left[ \frac{dy}{dx} \right]}_{x=a}}$
Thus,
$f'(a)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(a+h)-f(a)}{h}$

 Question 01

Find the derivative of:

$(i)y=\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})$
$(ii)y=\log \tan \left( \frac{x}{2}+\frac{\pi }{4} \right)$
$(iii)y={{10}^{{{10}^{x}}}}$
$(iv)y={{\sin }^{-1}}\sqrt{f(x)}$

Solution:
$y=\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})$
Differentiating both sides w.r.t. x we get,
$\frac{dy}{dx}=\frac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}.\frac{d}{dx}(x+\sqrt{{{x}^{2}}+{{a}^{2}}})$
$\Rightarrow \frac{dy}{dx}=\frac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}.\left\{ 1+\frac{2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}} \right\}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}.\left\{ \frac{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right\}$
$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}$

Solution:
$y=\log \tan \left( \frac{x}{2}+\frac{\pi }{4} \right)$
$\frac{dy}{dx}=\frac{1}{\tan \left( \frac{x}{2}+\frac{\pi }{4} \right)}.\frac{d}{dx}\tan \left( \frac{x}{2}+\frac{\pi }{4} \right)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\tan \left( \frac{x}{2}+\frac{\pi }{4} \right)}.{{\sec }^{2}}\left( \frac{x}{2}+\frac{\pi }{4} \right).\frac{d}{dx}\left( \frac{x}{2}+\frac{\pi }{4} \right)$
$\Rightarrow \frac{dy}{dx}=\frac{\cos \left( \frac{x}{2}+\frac{\pi }{4} \right)}{\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)}.\frac{1}{{{\cos }^{2}}\left( \frac{x}{2}+\frac{\pi }{4} \right)}.\frac{1}{2}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)\cos \left( \frac{x}{2}+\frac{\pi }{4} \right)}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sin 2\left( \frac{x}{2}+\frac{\pi }{4} \right)}$
$\therefore \frac{dy}{dx}=\frac{1}{\sin \left( x+\frac{\pi }{2} \right)}=\frac{1}{\cos x}=\sec x$

Solution:
$y={{10}^{{{10}^{x}}}}$
$\Rightarrow \frac{dy}{dx}={{10}^{{{10}^{x}}}}\log 10.\frac{d}{dx}({{10}^{x}})$
$\Rightarrow \frac{dy}{dx}={{10}^{{{10}^{x}}}}\log {{10.10}^{x}}.\log 10$
$\therefore \frac{dy}{dx}={{\left( \log 10 \right)}^{2}}{{10}^{{{10}^{x}}}}{{.10}^{x}}$

Solution:
$y={{\sin }^{-1}}\sqrt{f(x)}$
Differentiating both sides w.r.t. x we get,
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1+{{\left( \sqrt{f(x)} \right)}^{2}}}}.\frac{d}{dx}\sqrt{f(x)}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1+{{\left( \sqrt{f(x)} \right)}^{2}}}}.\frac{2}{2\sqrt{f(x)}}.\frac{d}{dx}f(x)$
$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1+{{\left( \sqrt{f(x)} \right)}^{2}}}}.\frac{{{f}^{‘}}(x)}{\sqrt{f(x)}}$

 Question 02

Find the derivative of:

$(i){{e}^{xy}}-4xy=4$
$(ii)x=a{{\cos }^{3}}\theta ,y=a{{\sin }^{3}}\theta$
$(iii)x={{\cos }^{-1}}\left( \frac{1-{{t}^{2}}}{1+{{t}^{2}}} \right),y={{\tan }^{-1}}\left( \frac{3t-{{t}^{3}}}{1-3{{t}^{2}}} \right)$
$(iv)x={{\cos }^{-1}}(8{{t}^{4}}-8{{t}^{2}}+1),y={{\sin }^{-1}}(3t-4{{t}^{3}})$

Solution:
${{e}^{xy}}-4xy=4$
${{e}^{xy}}\frac{d}{dx}(xy)-4\frac{d}{dx}(xy)=0$
$\Rightarrow {{e}^{xy}}\left( y+x\frac{dy}{dx} \right)-4\left( y+x\frac{dy}{dx} \right)=0$
$\Rightarrow \left( y+x\frac{dy}{dx} \right)({{e}^{xy}}-4)=0$
$\therefore \left( y+x\frac{dy}{dx} \right)=0,[\because ({{e}^{xy}}-4)\ne 0]$
$\therefore \frac{dy}{dx}=-\frac{y}{x}$

Solution:
$x=a{{\cos }^{3}}\theta ,y=a{{\sin }^{3}}\theta$
$x=a{{\cos }^{3}}\theta$
$\therefore \frac{dx}{d\theta }=a.3{{\cos }^{2}}\theta .(-\sin \theta )=-3a{{\cos }^{2}}\theta \sin \theta$
$y=a{{\sin }^{3}}\theta$
$\therefore \frac{dy}{d\theta }=a.3{{\sin }^{2}}\theta .\cos \theta =3a{{\sin }^{2}}\theta \cos \theta$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{{\sin }^{2}}\theta \cos \theta }{-3a{{\cos }^{2}}\theta \sin \theta }=-\tan \theta$

Solution:
$x={{\cos }^{-1}}\left( \frac{1-{{t}^{2}}}{1+{{t}^{2}}} \right),y={{\tan }^{-1}}\left( \frac{3t-{{t}^{3}}}{1-3{{t}^{2}}} \right)$
We know that,
${{\cos }^{-1}}\left( \frac{1-{{t}^{2}}}{1+{{t}^{2}}} \right)=2{{\tan }^{-1}}t$
${{\tan }^{-1}}\left( \frac{3t-{{t}^{3}}}{1-3{{t}^{2}}} \right)=3{{\tan }^{-1}}t$
Hence,
$x=2{{\tan }^{-1}}t,y=3{{\tan }^{-1}}t$
$\Rightarrow y=\frac{3}{2}x$
$\therefore \frac{dy}{dx}=\frac{3}{2}$

Solution:
$x={{\cos }^{-1}}(8{{t}^{4}}-8{{t}^{2}}+1),y={{\sin }^{-1}}(3t-4{{t}^{3}})$
Now,
$x={{\cos }^{-1}}(8{{t}^{4}}-8{{t}^{2}}+1)={{\cos }^{-1}}[2(4{{t}^{4}}-4{{t}^{2}})+1]$
$\Rightarrow x={{\cos }^{-1}}[2\{{{(2{{t}^{2}})}^{2}}-2.2{{t}^{2}}.1+1\}-1]$
$\Rightarrow x={{\cos }^{-1}}[2{{(2{{t}^{2}}-1)}^{2}}-1]$
$\Rightarrow x={{\cos }^{-1}}[2{{(2{{\cos }^{2}}\theta -1)}^{2}}-1],[let,t=\cos \theta ]$
$\Rightarrow x={{\cos }^{-1}}[2{{\cos }^{2}}2\theta -1]$
$\Rightarrow x={{\cos }^{-1}}\cos 4\theta =4\theta$
$\Rightarrow x=4{{\cos }^{-1}}t$
$\therefore \frac{dx}{dt}=\frac{4}{\sqrt{1-{{t}^{2}}}}$
And
$y={{\sin }^{-1}}(3t-4{{t}^{3}})=3{{\sin }^{-1}}t$
$\therefore \frac{dy}{dt}=\frac{3}{\sqrt{1-{{t}^{2}}}}$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{3}{\sqrt{1-{{t}^{2}}}}}{\frac{4}{\sqrt{1-{{t}^{2}}}}}=\frac{3}{4}$

 Question 03

Find the derivative of:
$(i)y={{\log }_{x}}\sin x+{{a}^{{{x}^{2}}}},(a>0)$
$(ii){{x}^{y}}={{e}^{x-y}}$
$(iii)y={{(\sin x)}^{\cos x}}+{{(\cos x)}^{\sin x}}$
$(iv)y={{\left( {{(\tan x)}^{\tan x}} \right)}^{\tan x}},at,x=\frac{\pi }{4}$
$(v)y={{x}^{{{e}^{x}}}}+{{e}^{{{x}^{x}}}}$
$(vi)y={{x}^{\log x}}+{{(\log x)}^{x}}$

Solution:
$y={{\log }_{x}}\sin x+{{a}^{{{x}^{2}}}}$
$\Rightarrow y=\frac{{{\log }_{e}}\sin x}{{{\log }_{e}}x}+{{a}^{{{x}^{2}}}}$
$\therefore \frac{dy}{dx}=\frac{{{\log }_{e}}x\frac{d}{dx}\left( {{\log }_{e}}\sin x \right)-{{\log }_{e}}\sin x\frac{d}{dx}({{\log }_{e}}x)}{{{\left( {{\log }_{e}}x \right)}^{2}}}+{{a}^{{{x}^{2}}}}\log a.\frac{d}{dx}({{x}^{2}})$
$\Rightarrow \frac{dy}{dx}=\frac{{{\log }_{e}}x.\frac{1}{\sin x}.\cos x-{{\log }_{e}}\sin x.\frac{1}{x}}{{{\left( {{\log }_{e}}x \right)}^{2}}}+{{a}^{{{x}^{2}}}}\log a.2x$
$\therefore \frac{dy}{dx}=\frac{x\cos x{{\log }_{e}}x-\sin x{{\log }_{e}}\sin x}{x\sin x{{\left( {{\log }_{e}}x \right)}^{2}}}+2x{{a}^{{{x}^{2}}}}\log a$

Solution:
${{x}^{y}}={{e}^{x-y}}$
Taking log at both sides we get,
$y\log x=x-y$
$\therefore y.\frac{1}{x}+\log x.\frac{dy}{dx}=1-\frac{dy}{dx}$
$\Rightarrow (1+\log x)\frac{dy}{dx}=1-\frac{y}{x}=\frac{x-y}{x}$
$\Rightarrow \frac{dy}{dx}=\frac{x-y}{x(1+\log x)}$
$\left[ \because y(1+\log x)=x,\therefore (1+\log x)=\frac{x}{y} \right]$
$\therefore \frac{dy}{dx}=\frac{x-y}{x.\frac{x}{y}}=\frac{y(x-y)}{{{x}^{2}}}$

Solution:
$y={{(\sin x)}^{\cos x}}+{{(\cos x)}^{\sin x}}$
We know that,
${{e}^{{{\log }_{e}}M}}=M$
$\therefore y={{e}^{{{\log }_{e}}{{(\sin x)}^{\cos x}}}}+{{e}^{{{\log }_{e}}{{(\cos x)}^{\sin x}}}}$
$\Rightarrow y={{e}^{\cos x{{\log }_{e}}(\sin x)}}+{{e}^{\sin x{{\log }_{e}}(\cos x)}}$
$\therefore \frac{dy}{dx}={{e}^{\cos x{{\log }_{e}}(\sin x)}}\frac{d}{dx}\left( \cos x{{\log }_{e}}(\sin x) \right)+{{e}^{\sin x{{\log }_{e}}(\cos x)}}\frac{d}{dx}\left( \sin x{{\log }_{e}}(\cos x) \right)$
$\Rightarrow \frac{dy}{dx}={{(\sin x)}^{\cos x}}\left[ \cos x.\frac{1}{\sin x}.\cos x+(-\sin x).{{\log }_{e}}(\sin x) \right]$
$+{{(\cos x)}^{\sin x}}\left[ \sin x.\frac{1}{\cos x}.(-\sin x)+\cos x.{{\log }_{e}}(\cos x) \right]$
$\therefore \frac{dy}{dx}={{(\sin x)}^{\cos x}}\left[ \cot x\cos x-\sin x{{\log }_{e}}(\sin x) \right]+{{(\cos x)}^{\sin x}}\left[ -\sin x\tan x+\cos x{{\log }_{e}}(\cos x) \right]$

Solution:
$y={{\left( {{(\tan x)}^{\tan x}} \right)}^{\tan x}}$
$\Rightarrow y={{\left( \tan x \right)}^{{{\tan }^{2}}x}}$
Taking log at both sides we get,
$\log y={{\tan }^{2}}x\log (\tan x)$
$\therefore \frac{1}{y}\frac{dy}{dx}={{\tan }^{2}}x.\frac{1}{\tan x}.{{\sec }^{2}}x+2\tan x.{{\sec }^{2}}x.\log (\tan x)$
$\Rightarrow \frac{dy}{dx}=y[\tan x{{\sec }^{2}}x+2\tan x.{{\sec }^{2}}x.\log (\tan x)]$
$\Rightarrow \frac{dy}{dx}={{\left( \tan x \right)}^{{{\tan }^{2}}x}}\tan x{{\sec }^{2}}x[1+2\log (\tan x)]$
$\Rightarrow {{\left[ \frac{dy}{dx} \right]}_{x=\frac{\pi }{4}}}={{\left( \tan \frac{\pi }{4} \right)}^{{{\tan }^{2}}\frac{\pi }{4}}}\tan \frac{\pi }{4}{{\sec }^{2}}\frac{\pi }{4}[1+2\log (\tan \frac{\pi }{4})]$
$\therefore {{\left[ \frac{dy}{dx} \right]}_{x=\frac{\pi }{4}}}=1.1.2[1+2.0]=2$

Solution:
$y={{x}^{{{e}^{x}}}}+{{e}^{{{x}^{x}}}}$
$\Rightarrow y={{e}^{\log {{x}^{{{e}^{x}}}}}}+{{e}^{\log {{e}^{{{x}^{x}}}}}}={{e}^{{{e}^{x}}\log x}}+{{e}^{{{x}^{x}}}}$
$\therefore \frac{dy}{dx}={{e}^{\log {{x}^{{{e}^{x}}}}}}\frac{d}{dx}\left( \log {{x}^{{{e}^{x}}}} \right)+{{e}^{{{x}^{x}}}}\frac{d}{dx}\left( {{x}^{x}} \right)$
$\Rightarrow \frac{dy}{dx}={{e}^{\log {{x}^{{{e}^{x}}}}}}\frac{d}{dx}\left( {{e}^{x}}\log x \right)+{{e}^{{{x}^{x}}}}\frac{d}{dx}\left( {{e}^{x\log x}} \right)$
$\Rightarrow \frac{dy}{dx}={{e}^{\log {{x}^{{{e}^{x}}}}}}\left[ {{e}^{x}}.\frac{1}{x}+{{e}^{x}}\log x \right]+{{e}^{{{x}^{x}}}}.{{e}^{x\log x}}\frac{d}{dx}\left( x\log x \right)$
$\Rightarrow \frac{dy}{dx}={{x}^{{{e}^{x}}}}.\frac{{{e}^{x}}}{x}[1+x\log x]+{{e}^{{{x}^{x}}}}.{{x}^{x}}\left( x.\frac{1}{x}+\log x \right)$
$\therefore \frac{dy}{dx}={{x}^{{{e}^{x}}-1}}{{e}^{x}}(1+x\log x)+{{e}^{{{x}^{x}}}}{{x}^{x}}\left( 1+\log x \right)$

Solution:
$y={{x}^{\log x}}+{{(\log x)}^{x}}={{e}^{\log x.\log x}}+{{e}^{x\log (\log x)}}$
$\Rightarrow y={{e}^{{{\left( \log x \right)}^{2}}}}+{{e}^{x\log (\log x)}}$
$\therefore \frac{dy}{dx}={{e}^{{{\left( \log x \right)}^{2}}}}\frac{d}{dx}\left( {{\left( \log x \right)}^{2}} \right)+{{e}^{x\log (\log x)}}\frac{d}{dx}\left( x\log (\log x) \right)$
$\Rightarrow \frac{dy}{dx}={{e}^{{{\left( \log x \right)}^{2}}}}.2\log x.\frac{1}{x}+{{e}^{x\log (\log x)}}\left[ x.\frac{1}{\log x}.\frac{1}{x}+\log (\log x).1 \right]$
$\Rightarrow \frac{dy}{dx}={{x}^{\log x}}.\frac{1}{x}.2\log x+{{(\log x)}^{x}}\left[ \frac{1}{\log x}+\log (\log x) \right]$
$\therefore \frac{dy}{dx}=2{{x}^{\log x-1}}.\log x+{{(\log x)}^{x}}\left[ \frac{1}{\log x}+\log (\log x) \right]$

 Question 04

Find the derivative of:
$(i){{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x},w.r.t.,{{\tan }^{-1}}x$
$(ii){{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}},w.r.t.,{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}$
$(iii){{\tan }^{-1}}\frac{t}{\sqrt{1-{{t}^{2}}}},w.r.t.,{{\sec }^{-1}}\frac{1}{\sqrt{2{{t}^{2}}-1}}$

Solution:
$let,y={{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x},and,z={{\tan }^{-1}}x\Rightarrow x=\tan z$
$\therefore y={{\tan }^{-1}}\frac{\sqrt{1+{{\tan }^{2}}z}-1}{\tan z}={{\tan }^{-1}}\frac{\sec z-1}{\tan z}$
$\Rightarrow y={{\tan }^{-1}}\frac{1-\cos z}{\sin z}={{\tan }^{-1}}\frac{2{{\sin }^{2}}\frac{z}{2}}{2\sin \frac{z}{2}\cos \frac{z}{2}}$
$\Rightarrow y={{\tan }^{-1}}\left( \tan \frac{z}{2} \right)=\frac{z}{2}$
$\therefore \frac{dy}{dz}=\frac{1}{2}$
This is the required derivative.

Solution:
$Let,y={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}},and,z={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}$
$\Rightarrow y=2{{\tan }^{-1}}x,and,z=2{{\tan }^{-1}}x$
$\Rightarrow y=z,\therefore \frac{dy}{dz}=1$
This is the required derivative.

Solution:
$Let,y={{\tan }^{-1}}\frac{t}{\sqrt{1-{{t}^{2}}}},and,z={{\sec }^{-1}}\frac{1}{\sqrt{2{{t}^{2}}-1}}$
Again let,
$t=\cos \theta$
$\therefore y={{\tan }^{-1}}\frac{\cos \theta }{\sqrt{1-{{\cos }^{2}}\theta }}={{\tan }^{-1}}\frac{\cos \theta }{\sin \theta }$
$\Rightarrow y={{\tan }^{-1}}\left( \cot \theta \right)={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{2}-\theta \right) \right)$
$\Rightarrow y=\frac{\pi }{2}-\theta ,\therefore \frac{dy}{d\theta }=-1$
And
$z={{\sec }^{-1}}\frac{1}{\sqrt{2{{\cos }^{2}}\theta -1}}={{\sec }^{-1}}\frac{1}{\sqrt{\cos 2\theta }}$
$\Rightarrow z={{\cos }^{-1}}\left( \sqrt{\cos 2\theta } \right)$
$\therefore \frac{dz}{d\theta }=-\frac{1}{\sqrt{1-\cos 2\theta }}.\frac{d}{d\theta }(\sqrt{\cos 2\theta })$
$\Rightarrow \frac{dz}{d\theta }=-\frac{1}{\sqrt{2{{\sin }^{2}}\theta }}.\frac{1}{2\sqrt{\cos 2\theta }}.(-2\sin 2\theta )$
$\Rightarrow \frac{dz}{d\theta }=-\frac{2\sin \theta \cos \theta }{\sqrt{2}\sin \theta .\sqrt{2{{\cos }^{2}}\theta -1}}=\frac{\sqrt{2}\cos \theta }{\sqrt{2{{\cos }^{2}}\theta -1}}$
$\therefore \frac{dz}{d\theta }=\frac{\sqrt{2}t}{\sqrt{2{{t}^{2}}-1}}$
$\therefore \frac{dy}{dz}=\frac{\frac{dy}{d\theta }}{\frac{dz}{d\theta }}=-1.\frac{\sqrt{2{{t}^{2}}-1}}{\sqrt{2}t}=-\sqrt{1-\frac{1}{2{{t}^{2}}}}$
This is the required derivative.

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