Question

If $x^y=e^{(x-y)},$ then find $dy/dx$

• A. $\log x/{(1+\log x)}^2$
• B. $(\log x)/(1+\log x)$
• C. $(1-\log x)/(1+\log x)$
• D. $(\log x)/{(1-\log x)}^2$

Answer 1

Answer: (A)

We have, $x^y=e^{(x-y)}$ Taking log on both sides, we get $y\log x=(x-y)\log e$
$\Rightarrow$ $y\log x=x-y$ ?.(i) On differentiating w. r. t. x, we get $y.\frac{1}{x}+\log x.\frac{dy}{dx}=1-\frac{dy}{dx}$
$\Rightarrow$ $\frac{dy}{dx}(\log x+1)=1-\frac{y}{x}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{x-y}{(\log x+1)x}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{x-\frac{x}{\log x+1}}{x(\log x+1)}$ $\left(fromEq.(i),y=\frac{x}{\log x+1}\right)$
$\Rightarrow$ $\frac{dy}{dx}=\frac{x\log x+x-x}{x{(\log x+1)}^2}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{x\log x}{x{(\log x+1)}^2}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{\log x}{{(1+\log x)}^2}$