Question

Differentiate $(\cos x)^{\sin x}$ w. r. t. $(\sin x)^{\cos x}$

Answer 1

very good questions.

Answer 2

Answer:

$\text{Consider }$

$y=u^v$

$\text{Take logarithm on both sides}$

$logy=log\,u^v$

$logy=v\,log\,u$

$\text{Differrentiate with respect to x}$

$\frac{1}{y}\frac{dy}{dx}=v(\frac{1}{u})\frac{du}{dx}+log\,u\:\frac{dv}{dx}$

$\frac{dy}{dx}=y[\frac{v}{u}\frac{du}{dx}+log\,u\:\frac{dv}{dx}]$

$\implies\:\boxed{\frac{d(u^v)}{dx}=u^v[\frac{v}{u}\frac{du}{dx}+log\,u\:\frac{dv}{dx}]}$..........(1)

Using (1)

$\frac{d((cosx)^{sinx})}{dx}=(cosx)^{sinx}[\frac{sinx}{cosx}\:\frac{d(cosx)}{dx}+log(cosx)\:\frac{d(sinx)}{dx}]$

$\frac{d((cosx)^{sinx})}{dx}=(cosx)^{sinx}[tanx\,(-sinx)+log(cosx)\,cosx]$

$\frac{d((cosx)^{sinx})}{dx}=(cosx)^{sinx}[-tanx\,sinx+cosx\,log(cosx)]$

Using (1)

$\frac{d((sinx)^{cosx})}{dx}=(sinx)^{cosx}[\frac{cosx}{sinx}\frac{d(sinx)}{dx}+log\,(sinx)\:\frac{d(cosx)}{dx}]$

$\frac{d((sinx)^{cosx})}{dx}=(sinx)^{cosx}[cotx\,(cosx)+log\,(sinx)\:(-sinx)]$

$\implies\:\frac{d((sinx)^{cosx})}{dx}=(sinx)^{cosx}[cotx\,cosx-sinx\,log\,(sinx)]$

$\text{Now }$

$\displaystyle\frac{d((cosx)^{sinx})}{d((cosx)^{sinx})}$

$=\displaystyle\frac{\frac{d((cosx)^{sinx})}{dx}}{\frac{d((cosx)^{sinx})}{dx}}$

$=\displaystyle\frac{(cosx)^{sinx}[-tanx\,sinx+cosx\,log(cosx)]}{(sinx)^{cosx}[cotx\,cosx-sinx\,log\,(sinx)]}$