Question

#Quality Question
@Derivatives

$\bf if \: x {}^{y} = e {}^{x - y} \\ \\ \bf then \: prove \: \: \dfrac{dy}{dx} = \frac{logx}{( {1 + logx)}^{2} }$
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Answer 1

▪Question :-

If  $\large \bf {x}^{y}  =  {e}^{x - y}$

Then Prove That :-

$\bf \dfrac{dy}{dx}  =  \dfrac{log \: x}{(1 +  {log \: x)}^{2} }$

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▪Given :-

$\Large \sf  {x}^{y}  = e {}^{x - y}$

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▪To Prove ➽

$\bf  \pink{  \dfrac{dy}{dx} =  \dfrac{log \: x}{(1 +  {log \: x)}^{2} }   }$

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▪ Formulae Used :-

$\bf \maltese \:  \:  \: log \:  {m}^{n}  = n .log \: m \\  \\ \small  \bf \maltese \:  \:  \:  \frac{d}{dx}  \bigg( \frac{f(x)}{g(x)}  \bigg) =  \frac{g(x).f '(x) - f(x).g '(x) }{ \{g(x) { \}}^{2} }  \\  \\  \bf \maltese \:  \:  \: log \: e = 1$

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▪ Proof :-

Given that,

$\large\sf{x}^{y}  =  {e}^{x - y}$

Taking log both side we get ,

$\sf log( {x}^{y} )  =  log( {e}^{x - y} )  \\  \\  \sf y. log \: x = (x - y). log \: e  \\  \\  \sf y.log \: x= x - y \\  \\  \sf y \big(log \: x + 1 \big) = x \\  \\   \large \purple{\bf \implies \underline{ \boxed{  \bf y =  \frac{x}{1 +  log \: x} }}}$

Differentiating both sides w.r.t x

$\sf \dfrac{dy}{dx}  =  \dfrac{d}{dx}  \bigg(  \dfrac{x}{1 + log \: x} \bigg) \\  \\  \sf =  \dfrac{(1 + log \: x) \frac{dx}{dx} - x \frac{d}{dx} (log \: x) }{ {(1 + log \: x)}^{2} }  \\  \\   \sf =  \frac{(1 + log \: x).1 - x. \frac{1}{x} }{ {(1 + log \: x)}^{2} }  \\  \\  =  \sf  \dfrac{1 + log \: x - 1}{ {(1 +  log \: x)}^{2} }  \\  \\ \Large \purple{ \implies  \underline {\boxed{{\bf  \frac{dy}{dx}  =  \frac{log \: x}{( {1 + log \: x)}^{2} }} }}}$

《Hence Proved 》

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Answer 2

$\Large{\underline{\underline{\bold{✯Given:}}}}$

• $\bf{x^{y} = e^{(x-y)}}$

$\Large{\underline{\underline{\red{\bold{➸\;To\;Prove:}}}}}$

• $\bf{\dfrac{d_y}{d_x} = \frac{logx}{( {1 + logx)}^{2} }}$

$\Large{\underline{\underline{\bold{\color{salmon}๛\; Required\; Response:}}}}$

$\bf{x^{y} = e^{(x-y)}}$

By Taking $\bold{log}$ on both sides

$\bf{log\;(x^{y}) = log\;(e^{x-y})}$

☮$\pink{\bold{\;log\;(a^{m}) = m\;.log\;a}}$

$\;\;\;\;\;\;\;\sf{y\;log\;x = (x-y)\;.log\;e}$

☮$\pink{\bold{\;log\;e = 1}}$

$\;\;\;\;\;\;\sf{y\;(log\;x) = x-y}$

$\;\;\;\;\;\;\sf{y+y\;log\:x = x}$

$\;\;\;\;\;\;\sf{y(1+log\;x) = x}$

$\;\;\;\;\;\;\;→\;\Large\bold{y = \frac{x}{1+log\;x}}$

On differentiating both sides with respect to x.

$→\;\Large{\bf{\frac{d_y}{d_x} = \frac{(1+log\;x)1-x(\frac{1}{x})}{(1+log\;x)²}}}$

$→\;\Large{\bf{\frac{d_y}{d_x} = \frac{(1+log\;x)-[\cancel{x}×\frac{1}{\cancel{x}}]}{(1+log\;x)²}}}$

$→\;\Large{\bf{\frac{d_y}{d_x} = \frac{1+log\;x-1}{(1+log\;x)²}}}$

$→\;\Large{\bf{\frac{d_y}{d_x} = \frac{log\;x}{(1+log\:x)²}}}$

Hence Proved,

• $\huge{\blue{\bold{\dfrac{d_y}{d_x} = \frac{log\;x}{( {1 + logx)}^{2} }}}}$

$\boxed{\tt{@MrSovereign}}$

Hope This Helps!!