Derivative of General Logarithm Function

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Theorem

Let $a \in \R_{>0}$ such that $a \ne 1$

Let $\log_a x$ be the logarithm function to base $a$.

Then:

$\map {\dfrac \d {\d x} } {\log_a x} = \dfrac {\log_a e} x$


Proof

\(\ds \map {\dfrac \d {\d x} } {\log_a x}\) \(=\) \(\ds \map {\dfrac \d {\d x} } {\dfrac {\log_e x} {\log_e a} }\) Change of Base of Logarithm
\(\ds \) \(=\) \(\ds \dfrac 1 {\log_e a} \map {\dfrac \d {\d x} } {\log_e x}\) Derivative of Constant Multiple
\(\ds \) \(=\) \(\ds \dfrac 1 {\log_e a} \dfrac 1 x\) Derivative of Natural Logarithm
\(\ds \) \(=\) \(\ds \dfrac {\log_a e} x\) Logarithm of Reciprocal

$\blacksquare$


Also presented as

This result can also be seen presented as:

$\map {\dfrac \d {\d x} } {\log_a x} = \dfrac 1 {x \ln a}$

where $\ln a := \log_e a$ is the natural logarithm of $a$.

This is seen to be equivalent to the given form by Logarithm of Reciprocal:

$\dfrac 1 {\ln a} = \log_a e$


Sources