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

The Correct Option is C

$ {{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 $