Question

If $x^y=e^{x-y},\mathrm{then}\frac{dy}{dx}$ is
(a) $\frac{1+x}{1+\log x}$

(b) $\frac{1-\log x}{1+\log x}$

(c) not defined

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

Answer 1

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

$$\begin{gathered} \mathrm{We}\mathrm{have},x^y=e^{x-y} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ \Rightarrow y\log x=\left(x-y\right){\log }_{e}e \\ \Rightarrow y\log x=x-y \\ \Rightarrow y\log x+y=x \\ \Rightarrow y\left(1+\log x\right)=x \\ \Rightarrow y=\frac{x}{1+\log x} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{\left(1+\log x\right)\times 1-x\times \left(0+{\displaystyle \frac{1}{x}}\right)}{{\left(1+\log x\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{1+\log x-1}{{\left(1+\log x\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{\log x}{{\left(1+\log x\right)}^2} \end{gathered}$$