If $$y=x^{sin\: x}+(log\:x)^x,\: find\: \large\frac{dy}{dx}.$$

Toolbox:
• $\log m^{\large n}=n\log m$
• $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x} • \large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} Step 1: y=x^{\sin x}+(\log x)^x Let u=x^{\large \sin x} Take \log on both sides \log u=\sin x\log x Differentiating w.r.t x we get, \large\frac{1}{u}\frac{du}{dx}$$=\sin x\large\frac{1}{x}$$+\log x.\cos x \qquad=\large\frac{\sin x}{x}$$+\cos x.\log x$
$\therefore \large\frac{du}{dx}$$=u[\large\frac{\sin x}{x}$$+\cos x.\log x]$
$\large\frac{du}{dx}$$=x^{\large\sin x}[\large\frac{\sin x}{x}$$+\cos x.\log x]$
Step 2:
Let $v=(\log x)^x$
Take $\log$ on both sides
$\log v=x\log(\log x)$
Differentiate w.r.t $x$ we get,
$\large\frac{1}{v}\frac{dv}{dx}$$=x.\large\frac{1}{\log x}.\frac{1}{x}$$+\log(\log x).1$
$\large\frac{dv}{dx}$$=v[\large\frac{1}{\log x}+$$\log(\log x)]$
$\large\frac{dv}{dx}$$=(\log x)^x[\large\frac{1}{\log x}+$$\log(\log x)]$
Step 3:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\quad\;\;\;=x^{\large\sin x}[\large\frac{\sin x}{x}$$+\cos x.\log x]+(\log x)^x[\large\frac{1}{\log x}+$$\log(\log x)]$
answered Sep 26, 2013