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{d y}{d x}+\frac{x}{y} \frac{d y}{d x}+\log y=0 $
$\left(\frac{x}{y}+\log x\right) \frac{d y}{d x}=-\left(\frac{y}{x}+\log y\right)$
$\frac{d y}{d x}=-\frac{y}{x} \frac{(y+x \log y)}{(x+y \log x)} $
$\left(\frac{d y}{d x}\right)_{at(2,2)}=\frac{-2}{2}\left(\frac{2+2 \log 2}{2+2 \log 2}\right)$
$=-1$