Question

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

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

Answer 1

Answer: (D)

Given $x^x=y^y$, taking log on the both sides, we get
$log{x}^x=log{y}^y$
$\Rightarrow xlogx=ylogy$,
Differentiating $w$.$r$.$t$. $x$, we get
$x\left(\frac{1}{x}\right)+logx\cdot 1=y\left(\frac{1}{y}\frac{dy}{dx}\right)+\left(logy\right)\frac{dy}{dx}$
$\Rightarrow 1+logx=\left(1+logy\right)\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1+logx}{1+logy}$.