# First Order Derivative Solution – Part I

Differential Calculus / Sunday, August 19th, 2018

# 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.

;

## 3 thoughts on “First Order Derivative Solution – Part I”

1. Rinku

I need some more
Thanks

2. essay of discipline

Thank you for another informative site. Where else may just I get that
kind of info written in such a perfect method? I’ve a project that I am
just now working on, and I’ve been on the look out for
such information.

3. Online casino

Excellent post. I was checking constantly this weblog and I am impressed!
Extremely helpful info particularly the remaining part 🙂
I take care of such information a lot. I was looking for this certain info for a
long time. Thank you and best of luck.