# Is this a valid way of taking the derivative.

$\def\lnx{\ln x}\def\lny{\ln y}$ 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?

• Yes. ${ } { }$ Nov 29 '14 at 20:31
• Youre a calculus student. Right? Are you unaware at this point that a single problem can be solved in several different ways? Nov 29 '14 at 22:51
• I asked because 1, I wasn't sure if all steps were correct. Reason why I was unsure was because points were taken off for solving this problem this way on an exam. Nov 30 '14 at 4:06

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