Question

$x^y.y^x=16$. Find $\frac{dy}{dx}$

Answer 1

$x^y.y^x=16,$ Find $\frac{dy}{dx}$

$\therefore$ taking log both side

$\therefore log(x^y.y^x)=log16$

$\therefore log{x}^y+log{y}^x=log16$

$\therefore ylogx+xlogy=log16$

$\therefore y.\frac{1}{x}+logx.\frac{dy}{dx}+x.\frac{1}{y}\frac{dy}{dx}+logy=0$

$\therefore \frac{dy}{dx}(logx+x/y)=-log-y/x$

$\therefore \frac{dy}{dx}=-\frac{(logy+y/x)}{(logx+x/y)}$