First Order Derivative Solution

This is the Part II of the First Order Derivative Solution series, if you not seen the First Order Derivative Solution Part I, I suggest you to check that. It will be beneficial for your respective exam preparation.

If you seen the Part I, then you ready to go, so let’s start the Part II of the series.

\[f(x)={{\left( \frac{a+x}{b+x} \right)}^{a+b+2x}}then\]
\[show,f'(0)=\left[ 2\log \left( \frac{a}{b} \right)+\frac{{{b}^{2}}-{{a}^{2}}}{ab} \right]{{\left( \frac{a}{b} \right)}^{a+b}}\]
Solution:
\[f(x)={{\left( \frac{a+x}{b+x} \right)}^{a+b+2x}}\]
\[\therefore \log f(x)=(a+b+2x).\log \left( \frac{a+x}{b+x} \right)\]
\[\therefore \frac{1}{f(x)}.f'(x)=2.\log \left( \frac{a+x}{b+x} \right)+(a+b+2x).\frac{b+x}{a+x}.\frac{(b+x).1-(a+x).1}{{{(b+x)}^{2}}}\]
\[\Rightarrow f'(x)=f(x)\left[ 2.\log \left( \frac{a+x}{b+x} \right)+\frac{(a+b+2x)(b-a)}{(a+x)(b+x)} \right]\]
\[\Rightarrow f'(x)={{\left( \frac{a+x}{b+x} \right)}^{a+b+2x}}\left[ 2.\log \left( \frac{a+x}{b+x} \right)+\frac{(a+b+2x)(b-a)}{(a+x)(b+x)} \right]\]
\[\Rightarrow f'(0)={{\left( \frac{a}{b} \right)}^{a+b}}\left[ 2.\log \left( \frac{a}{b} \right)+\frac{(a+b)(b-a)}{ab} \right]\]
\[\therefore f'(0)=\left[ 2\log \left( \frac{a}{b} \right)+\frac{{{b}^{2}}-{{a}^{2}}}{ab} \right]{{\left( \frac{a}{b} \right)}^{a+b}}\]

\[{{y}^{x}}={{e}^{y-x}}then-show,\frac{dy}{dx}=\frac{{{\left( \log ey \right)}^{2}}}{\log y}\]
Solution:
Given,
\[{{y}^{x}}={{e}^{y-x}}\]
\[\Rightarrow x\log y=y-x…….(1)\]
\[\therefore x.\frac{1}{y}.\frac{dy}{dx}+1.\log y=\frac{dy}{dx}-1\]
\[\Rightarrow \left( 1-\frac{x}{y} \right)\frac{dy}{dx}=1+\log y=\log e+\log y\]
\[\Rightarrow \frac{dy}{dx}=\frac{\log ey}{\left( 1-\frac{x}{y} \right)}=\frac{y\log ey}{y-x}=\frac{y\log ey}{x\log y}[from,(1)]\]
\[\left[ \because x(1+\log y)=y,\therefore \frac{y}{x}=\log ey \right]\]
\[\therefore \frac{dy}{dx}=\frac{y}{x}.\frac{\log ey}{\log y}=\frac{{{\left( \log ey \right)}^{2}}}{\log y}\]

\[(i)If,\cos y=x\cos (a+y)then-show\]
\[\frac{dy}{dx}=\frac{{{\cos }^{2}}(a+y)}{\sin a}\]
Solution:
Given,
\[\cos y=x\cos (a+y)\]
\[\Rightarrow x=\frac{\cos y}{\cos (a+y)}\]
\[\therefore \frac{dx}{dy}=\frac{\cos (a+y)(-\sin y)-\cos y(-\sin (a+y))}{{{\cos }^{2}}(a+y)}\]
\[\Rightarrow \frac{dx}{dy}=\frac{-\cos (a+y)\sin y+\cos y\sin (a+y)}{{{\cos }^{2}}(a+y)}\]
\[\Rightarrow \frac{dx}{dy}=\frac{sin(a+y-y)}{{{\cos }^{2}}(a+y)}=\frac{sina}{{{\cos }^{2}}(a+y)}\]
\[\therefore \frac{dy}{dx}=\frac{{{\cos }^{2}}(a+y)}{\sin a}\]

\[If,\sin y=x\sin (a+y),then-show\]
\[\frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+{{x}^{2}}}\]
Solution:
Given,
\[\sin y=x\sin (a+y)=x(\sin a\cos y+\cos a\sin y)\]
\[\Rightarrow (1-x\cos a)\sin y=x\sin a\cos y\]
\[\Rightarrow \tan y=\frac{x\sin a}{(1-x\cos a)}\]
\[\therefore {{\sec }^{2}}y\frac{dy}{dx}=\frac{(1-x\cos a).\sin a-x\sin a(-\cos a)}{{{(1-x\cos a)}^{2}}}\]
\[\Rightarrow (1+{{\tan }^{2}}y)\frac{dy}{dx}=\frac{\sin a}{{{(1-x\cos a)}^{2}}}\]
\[\Rightarrow \frac{dy}{dx}=\frac{\sin a}{(1+{{\tan }^{2}}y){{(1-x\cos a)}^{2}}}\]
\[\Rightarrow \frac{dy}{dx}=\frac{\sin a}{\left( 1+\frac{{{x}^{2}}{{\sin }^{2}}a}{{{(1-x\cos a)}^{2}}} \right){{(1-x\cos a)}^{2}}}\]
\[\Rightarrow \frac{dy}{dx}=\frac{\sin a}{{{(1-x\cos a)}^{2}}+{{x}^{2}}{{\sin }^{2}}a}\]
\[\Rightarrow \frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+{{x}^{2}}{{\cos }^{2}}a+{{x}^{2}}{{\sin }^{2}}a}\]
\[\therefore \frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+{{x}^{2}}}\]

\[(i)If,x\sqrt{1+y}+y\sqrt{1+x}=0,then-show\]
\[\frac{dy}{dx}=-\frac{1}{{{(1+x)}^{2}}}\]
Solution:
Given,
\[x\sqrt{1+y}+y\sqrt{1+x}=0\]
\[\Rightarrow x\sqrt{1+y}=-y\sqrt{1+x}\]
\[\Rightarrow {{\left( x\sqrt{1+y} \right)}^{2}}={{\left( -y\sqrt{1+x} \right)}^{2}}\]
\[\Rightarrow {{x}^{2}}(1+y)={{y}^{2}}(1+x)\]
\[\Rightarrow {{x}^{2}}-{{y}^{2}}=-xy(x-y)\]
\[\Rightarrow (x+y)(x-y)=-xy(x-y)\]
\[\Rightarrow x+y=-xy,[\because x-y\ne 0]\]
\[\Rightarrow y(1+x)=-x\]
\[\Rightarrow y=-\frac{x}{1+x}\]
\[\therefore \frac{dy}{dx}=-\left[ \frac{(1+x).1-x.1}{{{(1+x)}^{2}}} \right]=-\frac{1}{{{(1+x)}^{2}}}\]

\[If,\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=a(x-y),then-show\]
\[\frac{dy}{dx}=\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}},(x\ne y)\]
Solution:
\[Let,x=\sin \theta ,y=\sin \phi \]
\[\therefore \sqrt{1-{{\sin }^{2}}\theta }+\sqrt{1-{{\sin }^{2}}\phi }=a(\sin \theta -\sin \phi )\]
\[\Rightarrow \cos \theta +\cos \phi =a(\sin \theta -\sin \phi )\]
\[\Rightarrow \frac{\cos \theta +\cos \phi }{\sin \theta -\sin \phi }=a\]
\[\Rightarrow \frac{2\cos \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}}{2\cos \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2}}=a\]
\[\Rightarrow \cot \frac{\theta -\phi }{2}=a\Rightarrow \theta -\phi =2{{\cot }^{-1}}a\]
\[\Rightarrow {{\sin }^{-1}}x-{{\sin }^{-1}}y=2{{\cot }^{-1}}a\]
Differentiating w.r.t. x we get,
\[\frac{1}{\sqrt{1-{{x}^{2}}}}-\frac{1}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}=0\]
\[\therefore \frac{dy}{dx}=\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}},(x\ne y)\]

\[If,\sqrt{1-{{x}^{2n}}}+\sqrt{1-{{y}^{2n}}}=a({{x}^{n}}-{{y}^{n}}),then-show\]
\[\frac{dy}{dx}=\frac{{{x}^{n-1}}\sqrt{1-{{y}^{2n}}}}{{{y}^{n-1}}\sqrt{1-{{x}^{2n}}}},(x\ne y)\]
Solution:
\[Let,{{x}^{n}}=\sin \theta ,{{y}^{n}}=\sin \phi \]
\[\therefore \sqrt{1-{{\sin }^{2}}\theta }+\sqrt{1-{{\sin }^{2}}\phi }=a(\sin \theta -\sin \phi )\]
Now, same like above example.

\[(iv)If{{\sin }^{-1}}\left( \frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}} \right)=k,then-show,\frac{dy}{dx}=\frac{y}{x}\]
Solution:
\[{{\sin }^{-1}}\left( \frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}} \right)=k\]
\[\Rightarrow \frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\sin k\]
\[\Rightarrow \frac{({{x}^{2}}-{{y}^{2}})+({{x}^{2}}+{{y}^{2}})}{({{x}^{2}}-{{y}^{2}})-({{x}^{2}}+{{y}^{2}})}=\frac{\sin k+1}{\sin k-1}=p(let)\]
\[\Rightarrow -\frac{{{x}^{2}}}{{{y}^{2}}}=p\Rightarrow {{x}^{2}}=-p{{y}^{2}}\]
\[\therefore 2x=-p.2y\frac{dy}{dx}\]
\[\Rightarrow \frac{dy}{dx}=-\frac{x}{py}=\frac{-x}{-\frac{{{x}^{2}}}{{{y}^{2}}}.y}=\frac{y}{x}\]

\[If\frac{{{x}^{2}}}{{{p}^{2}}+\lambda }+\frac{{{y}^{2}}}{{{q}^{2}}+\lambda }=1,then-show\]
\[\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{1}{{{p}^{2}}}\left( x+y\frac{dy}{dx} \right)\left( x-y\frac{dx}{dy} \right)=1\]
Solution:
\[\frac{{{x}^{2}}}{{{p}^{2}}+\lambda }+\frac{{{y}^{2}}}{{{q}^{2}}+\lambda }=1\]
\[\Rightarrow ({{q}^{2}}+\lambda ){{x}^{2}}+({{p}^{2}}+\lambda ){{y}^{2}}=({{p}^{2}}+\lambda )({{q}^{2}}+\lambda )\]
Differentiating we get,
\[2({{q}^{2}}+\lambda )x+2({{p}^{2}}+\lambda )y\frac{dy}{dx}=0\]
\[\Rightarrow \frac{dy}{dx}=-\frac{({{q}^{2}}+\lambda )}{({{p}^{2}}+\lambda )}.\frac{x}{y}\]
\[L.H.S.=\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{1}{{{p}^{2}}}\left( x+y\frac{dy}{dx} \right)\left( x-y\frac{dx}{dy} \right)\]
\[=\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{1}{{{p}^{2}}}\left( x+y.\left( -\frac{({{q}^{2}}+\lambda )}{({{p}^{2}}+\lambda )}.\frac{x}{y} \right) \right)\left( x-y.\left( -\frac{({{p}^{2}}+\lambda )}{({{q}^{2}}+\lambda )}.\frac{y}{x} \right) \right)\]
\[=\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{1}{{{p}^{2}}}\left( x-\frac{({{q}^{2}}+\lambda )}{({{p}^{2}}+\lambda )}.x \right)\left( \frac{({{q}^{2}}+\lambda ){{x}^{2}}+({{p}^{2}}+\lambda ){{y}^{2}}}{x({{q}^{2}}+\lambda )} \right)\]
\[=\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{1}{{{p}^{2}}}\left( \frac{{{p}^{2}}+\lambda -{{q}^{2}}-\lambda }{({{p}^{2}}+\lambda )} \right).x\times \frac{({{p}^{2}}+\lambda )({{q}^{2}}+\lambda )}{x({{q}^{2}}+\lambda )}\]
\[=\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{{{p}^{2}}-{{q}^{2}}}{{{p}^{2}}}=1=R.H.S.\]

\[If,y=\frac{x\sqrt{{{x}^{2}}-{{a}^{2}}}}{2}-\frac{{{a}^{2}}}{2}\log \left( x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right),show\]
\[that,\frac{dy}{dx}=\sqrt{{{x}^{2}}-{{a}^{2}}}\]
Solution:
\[y=\frac{x\sqrt{{{x}^{2}}-{{a}^{2}}}}{2}-\frac{{{a}^{2}}}{2}\log \left( x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right)\]
\[\therefore \frac{dy}{dx}=\frac{1}{2}\left\{ 1.\sqrt{{{x}^{2}}-{{a}^{2}}}+x.\frac{1}{2\sqrt{{{x}^{2}}-{{a}^{2}}}}.2x \right\}\]
\[-\frac{{{a}^{2}}}{2}.\frac{1}{x+\sqrt{{{x}^{2}}-{{a}^{2}}}}.\left\{ 1+\frac{1}{2\sqrt{{{x}^{2}}-{{a}^{2}}}}.2x \right\}\]
\[\Rightarrow \frac{dy}{dx}=\frac{1}{2}\left\{ \frac{{{x}^{2}}-{{a}^{2}}+{{x}^{2}}}{\sqrt{{{x}^{2}}-{{a}^{2}}}} \right\}-\frac{{{a}^{2}}}{2}.\frac{1}{x+\sqrt{{{x}^{2}}-{{a}^{2}}}}.\frac{x+\sqrt{{{x}^{2}}-{{a}^{2}}}}{\sqrt{{{x}^{2}}-{{a}^{2}}}}\]
\[\Rightarrow \frac{dy}{dx}=\frac{2{{x}^{2}}-{{a}^{2}}}{2\sqrt{{{x}^{2}}-{{a}^{2}}}}-\frac{{{a}^{2}}}{2\sqrt{{{x}^{2}}-{{a}^{2}}}}\]
\[\therefore \frac{dy}{dx}=\frac{2{{x}^{2}}-2{{a}^{2}}}{2\sqrt{{{x}^{2}}-{{a}^{2}}}}=\sqrt{{{x}^{2}}-{{a}^{2}}}\]

\[If,{{x}^{2}}+{{y}^{2}}=t+\frac{1}{t},{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\]
\[show-that,\frac{dy}{dx}=-\frac{1}{{{x}^{3}}y}\]
Solution:
\[{{x}^{2}}+{{y}^{2}}=t+\frac{1}{t},{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\]
\[{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\]
\[\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\]
\[\Rightarrow {{\left( t+\frac{1}{t} \right)}^{2}}-2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\]
\[\Rightarrow {{t}^{2}}+\frac{1}{{{t}^{2}}}+2-2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\]
\[\Rightarrow {{y}^{2}}=\frac{1}{{{x}^{2}}}\]
\[\therefore 2y\frac{dy}{dx}=-\frac{2}{{{x}^{3}}}\Rightarrow \frac{dy}{dx}=-\frac{1}{{{x}^{3}}y}\]

\[If,x={{e}^{\cos 2t}},y={{e}^{\sin 2t}}show-that\frac{dy}{dx}=-\frac{y{{\log }_{e}}x}{x{{\log }_{e}}y}\]
Solution:
\[x={{e}^{\cos 2t}}\Rightarrow {{\log }_{e}}x=\cos 2t\]
\[\therefore \frac{1}{x}.\frac{dx}{dt}=-2\sin 2t\]
\[\Rightarrow \frac{dx}{dt}=-2x\sin 2t\]
And,
\[y={{e}^{\sin 2t}}\Rightarrow {{\log }_{e}}y=\sin 2t\]
\[\therefore \frac{1}{y}.\frac{dy}{dt}=2\cos 2t\]
\[\Rightarrow \frac{dy}{dt}=2y\cos 2t\]
\[\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2y\cos 2t}{-2x\sin 2t}=-\frac{y{{\log }_{e}}x}{x{{\log }_{e}}y}\]

If you have any question to solve, don’t be shy, put it in the comment box or send me.

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