Question

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

• A. - 1
• B. 0.0
• C. 1
• D. None of these

Answer 1

The correct option is A - 1

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