Derivative of Function to Power of Function

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Theorem

Let $\map u x, \map v x$ be real functions which are differentiable on $\R$.


Then:

$\map {\dfrac \d {\d x} } {u^v} = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v$


Proof

\(\ds \map {\dfrac \d {\d x} } {u^v}\) \(=\) \(\ds \map {\dfrac \d {\d x} } {\map \exp {v \ln u} }\) Definition of Power to Real Number
\(\ds \) \(=\) \(\ds \map \exp {v \ln u} \map {\dfrac \d {\d x} } {v \ln u}\) Chain Rule for Derivatives and Derivative of Exponential Function
\(\ds \) \(=\) \(\ds \map \exp {v \ln u} \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + v \map {\dfrac \d {\d x} } {\ln u} }\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds u^v \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + \frac v u \map {\dfrac \d {\d x} } u}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v\) gathering terms

$\blacksquare$


Also see

$\map {\dfrac \d {\d x} } {x^n} = n x^{n-1}$


$\map {\dfrac \d {\d x} } {a^x} = a^x \ln a$


Sources