Question

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

• A. 1
• B. 2
• C. $-1$
• D. $-2$

Answer 1

Answer: (C)

$-1$

Answer 2

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