TITLE: Is this a valid way of taking the derivative.
QUESTION [5 upvotes]: $\def\lnx{\ln x}\def\lny{\ln y}$
The problem is  find $f'(x)$ of $f(x)=x^{2\lnx}$ 
Here's my approach: 
Let $$y=x^{2\lnx}$$
$$\lny=\lnx^{2\lnx}$$
$$\lny=2\lnx\cdot\lnx$$
$$\lny=2(\lnx)^{2}$$
$${d\over dx}\lny = {d\over dx}2(\lnx)^{2}$$
$${1\over y}*y' = 2*2lnx*{1\over x}$$
$$y'=y*{4\lnx\over x}$$
$$y'=x^{2\lnx}*{4\lnx\over x}$$
My professor did it by taking the $\ln$ of $x^{2\lnx}$ and then using base $e$ something like 
$$e^{\lnx^{2\lnx}}$$ 
Is my approach valid?

REPLY [3 votes]: Your approach is perfectly correct.  Note that one final simplification is possible; $$y' = 4 x^{-1 + 2 \log x} \log x.$$