Choose the correct alternative in the following:

Question:

Choose the correct alternative in the following:

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

A. $\frac{1+\mathrm{x}}{1+\log \mathrm{x}}$

B. $\frac{1-\log x}{1+\log x}$

C. not defined

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

Solution:

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

Taking log both sides we get

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

$y \log x=(x-y) \log e$

$y \log x=(x-y) \because \log e=1$

$y=\frac{x}{\log x+1}$

Differentiating w.r.t $x$ we get,

$\frac{d y}{d x}=\frac{1 \cdot(\log x+1)-\frac{1}{x} \cdot x}{(\log x+1)^{2}}$

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

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