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

Toolbox:
• $(uv)'=u'v+uv'$
• $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
$y=(\log x)^x+x^{\large\cos x}$
Let $u=(\log x)^x$
Take $\log$ on both sides,
$\log u=x\log (\log x)$
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}$$=x.\large\frac{1}{\log x}.\frac{1}{x}$$+\log (\log x).1$
$\qquad\quad=\large\frac{1}{\log x}$$+\log(\log x) \large\frac{du}{dx}=$$u.[\large\frac{1}{\log x}$$+\log(\log x)] \quad\;\;=$$(\log x)^x\large\frac{1}{\log x}$$+\log(\log x) Step 2: Let v=x^{\large\cos x} Take \log on both sides, \log v=\cos x.\log x Differentiating with respect to x on both sides we get, \large\frac{1}{v}\frac{dv}{dx}$$=\cos x.\large\frac{1}{x}$$+\log x(-\sin x) \qquad=\large\frac{\cos x}{ x}$$-\sin x\log x$
$\large\frac{dv}{dx}=$$v.[\large\frac{\cos x}{ x}$$-\sin x\log x]$
$\large\frac{dv}{dx}=$$x^{\large\cos x}.[\large\frac{\cos x}{ x}$$-\sin x\log x]$
Step 3:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\quad\;\;=(\log x)^x\large\frac{1}{\log x}$$+\log(\log x)+x^{\large\cos x}.[\large\frac{\cos x}{ x}$$-\sin x\log x]$