Question

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

• A. – 1
• B. 1
• C. none of these

Answer 1

The correct option is A – 1

$x^y\cdot y^x=16$
\Rightarrow y log_e x + x log_e y = 4 log_e 2\)
Now, differentiating both sides w.r.t. x
$\frac{y}{x}+lo{g}_{e}x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+lo{g}_{e}y.1=0\therefore \frac{dy}{dx}=-\frac{(lo{g}_{e}y+\frac{y}{x})}{(lo{g}_{e}x+\frac{y}{x})}\therefore \frac{dy}{dx}{|}_{(}2,2)=-\frac{(lo{g}_{e}2+1)}{(lo{g}_{e}2+1)}=-1$