Derivative of Function to Power of Function

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Theorem

Let $u \left({x}\right), v \left({x}\right)$ be real functions which are differentiable on $\R$.


Then:

$D_x \left({u^v}\right) = v u^{v-1} D_x \left({u}\right) + u^v \left({\ln u}\right) D_x \left({v}\right)$


Proof

\(\displaystyle D_x \left({u^v}\right)\) \(=\) \(\displaystyle D_x \left({ \exp \left({v \ln u}\right) }\right)\) Power to Real Number
\(\displaystyle \) \(=\) \(\displaystyle \exp \left({v \ln u}\right) D_x \left({v \ln u}\right)\) Chain Rule for Derivatives and Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle \exp \left({v \ln u}\right) \left({\left({\ln u}\right) D_x \left({v}\right) + v D_x \left({\ln u}\right)}\right)\) Product Rule
\(\displaystyle \) \(=\) \(\displaystyle u^v \left({\left({\ln u}\right) D_x \left({v}\right) + \frac v u D_x \left({u}\right)}\right)\) Chain Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle v u^{v-1} D_x \left({u}\right) + u^v \left({\ln u}\right) D_x \left({v}\right)\) gathering terms

$\blacksquare$


Also see

$D_x \left({x^n}\right) = n x^{n-1}$


$D_x \left({a^x}\right) = a^x \ln a$


Sources