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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.

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My answer here may be helpful: math.stackexchange.com/a/164810/409 –  Blue Dec 5 '12 at 5:12

3 Answers 3

up vote 4 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}

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

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Regla de la cadena (chain rule):

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

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

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