Question

Find $\frac{dy}{dx}$ of function
$(cosx)^y=(cosy)^x$

Answer 1

The correct answer is $∴\frac{dy}{dx}=\frac{ytanx+log\,cosy}{xtany+log\,cosx}$
The given function is $(cosx)^y=(cosy)^x$
Taking logarithm on both the sides,we obtain
$y\,log\,cosx=x\,log\,cosy$
Differentiating both sides with respect to $x$,we obtain
$log\,cosx.\frac{dy}{dx}+y.\frac{d}{dx}(log\,cosx)=log\,cosy.\frac{d}{dx}(x)+x.\frac{d}{dx}(log\,cosy)$
$⇒log\,cosx\frac{dy}{dx}+y.\frac{1}{cosx}.\frac{d}{dx}(cosx)=log\,cosy.1+x.\frac{1}{cosy}.\frac{d}{dx}(cosy)$
$⇒log\,cosx\frac{dy}{dx}+\frac{y}{cosx}.(-sinx)=log\,cosy+\frac{x}{cosy}(-siny).\frac{dy}{dx}$
$⇒log\,cosx\frac{dy}{dx}-ytanx=log\,cosy-xtany\frac{dy}{dx}$
$⇒(log\,cosx+xtany)\frac{dy}{dx}=ytanx+log\,cosy$
$∴\frac{dy}{dx}=\frac{ytanx+log\,cosy}{xtany+log\,cosx}$