Question

If $x^y{y}^x=16$, then $\frac{dy}{dx}$ at $(2,2)$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$
• E. $0$

Answer 1

Answer: (C)

$x^y{y}^x=16$
Taking log on both sides,
$y\log x+x\log y=\log 16$
Differentiating on both sides,
$\frac{y}{x}+\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\log y=0$
$\left(\frac{x}{y}+\log x\right)\frac{dy}{dx}=-\left(\frac{y}{x}+\log y\right)$
$\frac{dy}{dx}=-\frac{y}{x}\frac{(y+x\log y)}{(x+y\log x)}$
${\left(\frac{dy}{dx}\right)}_{at(2,2)}=\frac{-2}{2}\left(\frac{2+2\log 2}{2+2\log 2}\right)$
$=-1$