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

  • 1
    $\begingroup$ Yes. $ { } { } $ $\endgroup$ – Hakim Nov 29 '14 at 20:31
  • $\begingroup$ Youre a calculus student. Right? Are you unaware at this point that a single problem can be solved in several different ways? $\endgroup$ – CogitoErgoCogitoSum Nov 29 '14 at 22:51
  • $\begingroup$ 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. $\endgroup$ – Nolohice Nov 30 '14 at 4:06

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

  • $\begingroup$ Any idea why my professor wouldn't want me to do it this way? He took 2 points off the exam and said I should have done it his way. $\endgroup$ – Nolohice Nov 29 '14 at 20:54
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    $\begingroup$ I don't know. The two methods are mathematically equivalent, so unless there is a mathematical reason why the professor can justify that your approach is incorrect, I don't see why deducting points for your answer would be reasonable. $\endgroup$ – heropup Nov 29 '14 at 21:06
  • $\begingroup$ My guess is that the professor wants to encourage students to solve math problems involving more complicated powers in a standard way: always convert to base e. That is indeed good advice! However, in my opinion the professor should respect alternative methods such as yours. $\endgroup$ – M. Wind Nov 29 '14 at 21:52

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