Question

If $x^y=e^{x-y},\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{at}x=e$ is

• A. 1
• B. -1
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{1}{4}$

Given that
$x^y=e^{x-y}\mathrm{Taking}\log \mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{base}e,\mathrm{we}\mathrm{get}\Rightarrow y\log x=x-y\Rightarrow y\left(1+\log x\right)=x\Rightarrow y=\frac{x}{1+\log x}\mathrm{Differentiating}w.r.t.x\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(1+\log x\right)\cdot 1-x\cdot \frac{1}{x}}{{\left(1+\log x\right)}^2}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\log x-1}{{\left(1+\log x\right)}^2}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\log x}{{\left(1+\log x\right)}^2}\mathrm{At}x=e{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{x=e}=\frac{\log e}{{\left(1+\log e\right)}^2}=\frac{1}{4}$