Question

If $x^y=e^{x-y}$ then $\frac{dy}{dx}=$

• A. $\frac{\log x}{{\left(1+\log x\right)}^2}$
• B. $-\frac{\log x}{{\left(1+\log x\right)}^2}$
• C. $-\frac{\log x}{{\left(1-\log x\right)}^2}$
• D. $\frac{\log x}{{\left(1-\log x\right)}^2}$

Answer 1

The correct option is A $\frac{\log x}{{\left(1+\log x\right)}^2}$

Given $x^y=e^{x-y}$

Taking logarithm both sides, $\log x^y=\log e^{(x-y)}$ $\Rightarrow y\times \log x=(x-y)\times \log e$

$\Rightarrow y\log x=x-y\Rightarrow y+y\log x=x\Rightarrow y(1+\log x)=x\Rightarrow y=\frac{x}{1+\log x}$

$\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

Here, let $u=x$ and $v=1+\log x$

$\therefore \frac{d}{dx}\left(\frac{x}{1+\log x}\right)=\frac{(1+\log x)\frac{dx}{dx}-x\frac{d(1+\log x)}{dx}}{(1+\log x{)}^2}=\frac{1+\log x-x(0+\frac{1}{x})}{(1+\log x{)}^2}=\frac{\log x}{(1+\log x{)}^2}$