A) 1
B) 2
C) \[-1\]
D) \[-2\]
E) 0
Correct Answer: C
Solution :
\[{{x}^{y}}{{y}^{x}}=16\] Taking log on both sides, \[y\text{ }log\text{ }x+x\text{ }log\text{ }y=log\text{ }16\] Differentiating on both sides, \[\frac{y}{x}+\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\log y=0\] \[\left( \frac{x}{y}+\log x \right)\frac{dy}{dx}=-\left( \frac{y}{x}+\log y \right)\] \[\frac{dy}{dx}=-\frac{y}{x}\frac{(y+x\log y)}{(x+y\log x)}\] \[{{\left( \frac{dy}{dx} \right)}_{at(2,2)}}=\frac{-2}{2}\left( \frac{2+2\log 2}{2+2\log 2} \right)=-1\]
You need to login to perform this action.
You will be redirected in
3 sec
