Question

If x^y = e^x-y prove that dy/dx = logx/ (1+log)²​

Answer 1

☆ Appropriate Question ➽

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

Then Prove That :-

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

☆ Given ➽

$\Large \sf {x}^{y} = e {}^{x - y}$

☆ To Prove ➽

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

☆ Formulae Used ➽

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

☆ 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

Ok Here is your Answer in picture attached