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

Yo lo calculé con Wolfram Alpha, pero no entendí muy bien cómo hacerlos. Gracias por las recomendaciones.

I calculated it using Wolfram Alpha, but I didn't understand very well how to do them. Thanks for any advice.

share|cite|improve this question
My answer here may be helpful: – Blue Dec 5 '12 at 5:12
up vote 5 down vote accepted

\begin{align} y&=f(x)^{g(x)}\\ \ln y&=g(x)\ln f(x)\\ \dfrac{dy}{dx}\dfrac{1}{y}&=g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)}\\ \dfrac{dy}{dx}&=y[g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)}]\\ \dfrac{dy}{dx}&=f(x)^{g(x)}\left[g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)}\right]\\ \end{align}

share|cite|improve this answer

$$ \begin{align} \def\d{\frac{\mathrm d}{\mathrm dx}} \d\left(f(x)^{g(x)}\right) &= \d\left(\left(\mathrm e^{\log f(x)}\right)^{g(x)}\right) \\ &= \d\left(\mathrm e^{g(x)\log f(x)}\right) \\ &= \mathrm e^{g(x)\log f(x)}\d(g(x)\log f(x)) \\ &= f(x)^{g(x)}\left(g'(x)\log f(x)+\frac{g(x)f'(x)}{f(x)}\right)\;. \end{align} $$

share|cite|improve this answer

Regla de la cadena (chain rule):

Con una constante $k$ (with a constant $k$):

$$(1)\;\;\;\;\;\;\;\;\;\left(f(x)^k\right)'=kf(x)^{k-1}\,f'(x)$$ $$(2)\;\;\;\;\;\;\;\;\;k>0\Longrightarrow\left(k^{g(x)}\right)'=k^{g(x)}\log k\cdot g'(x)$$

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.