Derivative of Powers of Functions

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Theorem

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


Then:

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


Proof

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle D_x \left({u^v}\right)\) \(=\) \(\displaystyle D_x \left({e^{v \ln u} }\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Powers of Real Numbers          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle e^{v \ln u} D_x \left({v \ln u}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Chain Rule and Derivative of Exponential Function          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle e^{v \ln u} \left({\ln u D_x \left({v}\right) + v D_x \left({\ln u}\right)}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Product Rule          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle u^v \left({\ln u D_x \left({v}\right) + \frac v u D_x \left({u}\right)}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Chain Rule          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle v u^{v-1} D_x \left({u}\right) + u^v \ln u D_x \left({v}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          gathering terms          

$\blacksquare$


Note

When $u = x$ and $v = n$ where $n$ is constant, we get the Power Rule for Derivatives:

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


When $v = x$ and $u = a$ where $a$ is constant, we get the Derivative of Exponential Function:

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


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