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,
\[f'(a)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(a+h)-f(a)}{h}\]

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,
\[\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1+{{\left( \sqrt{f(x)} \right)}^{2}}}}.\frac{d}{dx}\sqrt{f(x)}\]
\[\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1+{{\left( \sqrt{f(x)} \right)}^{2}}}}.\frac{2}{2\sqrt{f(x)}}.\frac{d}{dx}f(x)\]
\[\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1+{{\left( \sqrt{f(x)} \right)}^{2}}}}.\frac{{{f}^{‘}}(x)}{\sqrt{f(x)}}\]

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
\[y={{\sin }^{-1}}(3t-4{{t}^{3}})=3{{\sin }^{-1}}t\]
\[\therefore \frac{dy}{dt}=\frac{3}{\sqrt{1-{{t}^{2}}}}\]
\[\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{3}{\sqrt{1-{{t}^{2}}}}}{\frac{4}{\sqrt{1-{{t}^{2}}}}}=\frac{3}{4}\]

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

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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