Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Simplified to a very basic problem, is there a standard procedure for these types of differentiations?

$$\displaylines{ y = {x^{{e^x}}} \cr {{d} \over {dx}}\left( y \right) = {{d} \over {dx}}\left( {{x^{{e^x}}}} \right) \cr = {x^{{e^x}}}\left( {{e^x}\ln \left( x \right) + {{{e^x}} \over x}} \right) \cr} $$

share|cite|improve this question
What you have is correct. The standard procedures you probably already know: power rule, product rule, quotient rule, chain rule, derivatives of the basic functions. Also, rewriting $x^{f(x)}=e^{f(x)\ln x}$ crops up sometimes. What else do you want? – anon Mar 20 '13 at 6:07
up vote 1 down vote accepted

Here is an approach.

Take logarithm on both sides, $$\log y=\log x^{e^x}=e^x\log x$$ Differentiate both sides, $$\frac{y'}{y}=e^x\log x+\frac{e^x}{x}$$ Multiply both sides by $y$ and substitute the original equation $$\frac{dy}{dx}=y'=(e^x\log x+\frac{e^x}{x})y=(e^x\log x+\frac{e^x}{x})x^{e^x}$$

share|cite|improve this answer
I like to leave something for OP to think about. – Gerry Myerson Mar 20 '13 at 6:07
@GerryMyerson I think this trick is taught in all high schools. Did you come up with this idea on your own? – NECing Mar 20 '13 at 6:08
The idea of logarithmic differentiation? No, I didn't come up with it on my own. But I might have worked it out if someone had told me, take logs & differentiate --- and I would have had a feeling of accomplishment if I had been able to work it out from that. – Gerry Myerson Mar 20 '13 at 6:11
@GerryMyerson I didn't post the actual problem I am working on for precisely that reason. Thanks. – Patrick Mar 20 '13 at 6:12

Hint: take logarithms on both sides, then differentiate, recalling $(\log y)'=y'/y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.