Question

If $x^x=y^y$ then $\frac{dy}{dx}$ is

• A. $-\frac{x}{y}$
• B. $-\frac{y}{x}$
• C. $\frac{1+\log x}{1+\log y}$
• D. $1+\log \left(\frac{x}{y}\right)$

Answer 1

Answer: (C)

Given, $x^x=y^y$
Taking log on both sides, we get
$x\log x=y\log y$
Differentiating w.r.t. y, we get
$y.\frac{1}{y}.\frac{dy}{dx}+\log y$
$\frac{dy}{dx}=x\frac{1}{x}+\log x$
$\Rightarrow \frac{dy}{dx}\left(1+\log y\right)$
$=1+\log x$
$\Rightarrow \frac{dy}{dx}=\frac{1+\log x}{1+\log y}$