Question

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

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

Answer 1

Answer: (E)

$\frac{y(x-y)}{x^2}$

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

Take natural logarithm both sides

$\Rightarrow y\ln x=x-y$....(1)

Differentiate both sides w.r.t. $x$

$\Rightarrow \frac{y}{x}+\ln x\frac{dy}{dx}=1-\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}(\ln x+1)=1-\frac{y}{x}=\frac{x-y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{x-y}{x(\ln x+1)}=\frac{x-y}{x(\frac{x-y}{y}+1)}$, use (1)

$=\frac{y(x-y)}{x^2}$