Question

If $x^{y}=e^{x-y}$, then $\frac{d y}{d x}$ 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, $x^{y}=e^{x-y}$
$\Rightarrow y \log x=(x-y) \log e=x-y$
$\Rightarrow y=\frac{x}{1+\log x}$
$\Rightarrow \frac{d y}{d x}=\frac{(1+\log x) \cdot 1-x \cdot \frac{1}{x}}{(1+\log x)^{2}}=\frac{\log x}{(1+\log x)^{2}}$