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