Question

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

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

Answer 1

Answer: (A)

$\frac{logx}{(1+logx{)}^2}$

According to Question,

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

Taking log on both sides, We get

$y\log x=x-y$

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

On differentiating w.r.t $x$, we get

$\frac{dy}{dx}=\frac{1+\log x-x.\frac{1}{x}}{(1+\log x{)}^2}$

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