Question

If $x^y=e^{x-y}$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

Answer: (A)

We have, $x^y=e^{x-y}$
$\Rightarrow y\log x=(x-y)\log e=x-y$
$\Rightarrow y=\frac{x}{1+\log x}$
$\Rightarrow \frac{dy}{dx}=\frac{(1+\log x)\cdot 1-x\cdot \frac{1}{x}}{(1+\log x{)}^2}=\frac{\log x}{(1+\log x{)}^2}$