Question

Find $\frac{dy}{dx}$of the functions given in question.

$(cosx{)}^y=cosy{)}^x$.

Answer 1

Given, $(cosx{)}^y=(cosy{)}^x$., taking log on both sides, we get

$\left\{\right(cosx{)}^y\}=log\{\left(cosy{)}^x\right\}orylog\left(cosx\right)=xlog\left(cosy\right)Differentiatingbothsidesw.r.t.x,wegety\frac{d}{dx}\left(logcosx\right)+logcosx\frac{d}{dx}y=x\frac{d}{dx}\left(logcosy\right)+logcosy\frac{d}{dx}x\left(Usingproductrule\right)y\left(\frac{1}{cosx}\right)(-sinx)+log\left(cosx\right)\frac{dy}{dx}=x\left(\frac{1}{cosy}\right)(-siny)\frac{dy}{dx}+log\left(cosy\right).1\Rightarrow log\left(coss\right)\frac{dy}{dx}+xtany\frac{dy}{dx}=log\left(cosy\right)+ytanx\Rightarrow \left(log\right(cosx)+xtany)\frac{dy}{dx}=log\left(cosy\right)+ytanx\Rightarrow \frac{dy}{dx}=\frac{log\left(cosy\right)+y\left(tanx\right)}{log\left(cosx\right)+xtany}$