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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation