Question

if x^y=e^x-y, then dy/dx at x=e is​

Answer 1

Given,

$\longrightarrow x^y=e^{x-y}$

Taking log on both sides,

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

$\longrightarrow y\log x=x-y$

$\longrightarrow y+y\log x=x$

$\longrightarrow y(1+\log x)=x$

$\longrightarrow y=\dfrac{x}{1+\log x}$

Differentiating with respect to $x,$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{x}{1+\log x}\right)$

By quotient rule,

$\longrightarrow \dfrac{dy}{dx}=\dfrac{(1+\log x)\cdot\dfrac{d}{dx}(x)-x\cdot\dfrac{d}{dx}(1+\log x)}{(1+\log x)^2}$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{(1+\log x)(1)-x\left(0+\dfrac{1}{x}\right)}{(1+\log x)^2}$

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

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

At $x=e,$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{\log e}{(1+\log e)^2}$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{1}{(1+1)^2}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=\dfrac{1}{4}}}$

Hence $\bf{\dfrac{1}{4}}$ is the answer.