# Differentiate the functions given in w.r.t. $x :$ $y=(\log x)^{\large x}+x^{\large\log x}$

This question has appeared in model paper 2012

Toolbox:
• $(uv)'=u'v+uv'$
• $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
Let $y=(\log x)^{\large x}+x^{\large\log x}$
Now $u=(\log x)^{\large x}$
Taking $\log$ on both sides
$\log u=\log(\log x)^{\large x}$
$\qquad=x\log(\log x)$
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}=$$1.\log(\log x)+x.\large\frac{d}{dx}$$\log(\log x)$
$\qquad\;=\log(\log x)+x.\large\frac{1}{\log x}\large\frac{d}{dx}$$(\log x) \qquad\;=\log(\log x)+x.\large\frac{1}{\log x}\large\frac{1}{ x} \large\frac{du}{dx}$$=u[\log(\log x)+\large\frac{1}{\log x}]$
$\;\;\;\;=(\log x)^{\large x}$$[\log(\log x)+\large\frac{1}{\log x}] Step 2: Now consider v v=x^{\large\log x} Taking \log on both sides \log v=\log x^{\large\log x} \qquad=\log x.\log x \qquad=(\log x)^2 Differentiating with respect to x \large\frac{1}{v}\frac{dv}{dx}=$$2\log x.\large\frac{d}{dx}$$(\log x) \qquad\;=2\log x.\large\frac{1}{x} \qquad\;=\large\frac{2}{x}$$\log x$
$\large\frac{dv}{dx}=$$v.[\large\frac{2}{x}$$\log x]$
Substitute the value of $v$
$\large\frac{dv}{dx}=$$x^{\large\log x}[\large\frac{2}{x}$$\log x]$
$\quad=\large\frac{2}{x}$$x^{\large\log x}$$\log x$
Step 3:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$