Question

Differentiate $\sin x^{\cos x}+\cos x^{\sin x}$ with respect to $x$.

Answer 1

Let $y=\sin x^{\cos x}+\cos x^{\sin x}=u+v$

$u=\sin x^{\cos x}$
$\log u=\cos x\log \sin x$
Differentiate w.r.t. $x$
$\frac{1}{u}\frac{du}{dx}=\cos x\frac{d}{dx}\log \sin x+\log \sin x\frac{d}{dx}\cos x$
$\frac{1}{u}\frac{du}{dx}=\cos x\times \frac{1}{\sin x}\times \cos x-\log \sin x\times \sin x$
$\frac{1}{u}\frac{du}{dx}=\cos x\times \cot x-\sin x\times \log \sin x$
$\frac{du}{dx}=\sin x^{\cos x}[\cos x\cdot \cot x-\sin x\cdot \log \sin x]$

$v=\cos x^{\sin x}$
$\log v=\sin x\times \log \cos x$
Differentiate w.r.t. $x$
$\frac{1}{v}\frac{dv}{dx}=\sin x\frac{d}{dx}\log \cos x+\log \cos x\times \frac{d}{dx}\sin x$
$\frac{1}{v}\frac{dv}{dx}=\sin x\times \frac{1}{\cos x}\times (-\sin x)+\log \cos x\times \cos x$
$\frac{1}{v}\frac{dv}{dx}=-\sin x\times \tan x+\cos x\times \log \cos x$
$\frac{dv}{dx}=\cos x^{\sin x}[-\sin x\cdot \tan x+\cos x\cdot \log \cos x]$

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\therefore \frac{dy}{dx}=\left\{\sin x^{\cos x}\right[\cos x\cdot \cot x-\sin x\cdot \log \sin x\left]\right\}+\left\{\cos x^{\sin x}\right[-\sin x\cdot \tan x+\cos x\cdot \log \cos x\left]\right\}$