If x^y=e^ x-y , then d y d x is - Vidyadip

MCQ

If $x^y=e^{x-y}$, then $\frac{d y}{d x}$ is

A
$\frac{1+x}{1+\log x}$
B
$\frac{1-\log x}{1+\log x}$
✓
$\frac{\log x}{(1+\log x)^2}$
D
not defined

✓

Answer

Correct option: C.

$\frac{\log x}{(1+\log x)^2}$

(c) : We have, $x^y=e^{x-y}$
Taking $\log$ on both sides, we get
$
y \log x=(x-y) \log e \Rightarrow y=\frac{x}{1+\log x}
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{(1+\log x)-x \cdot \frac{1}{x}}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}
$

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