Nth Derivative of Natural Logarithm
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Theorem
The $n$th derivative of $\map \ln x$ for $n \ge 1$ is:
- $\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$
Proof
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
- $\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$
Basis for the Induction
$\map P 1$ is true, as this just says:
- $\dfrac \d {\d x} \ln x = \dfrac 1 x$
This follows by Derivative of Natural Logarithm Function.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\dfrac {\d^k} {\d x^k} \ln x = \dfrac {\paren {k - 1}! \paren {-1}^{k - 1} } {x^k}$
Then we need to show:
- $\dfrac {\d^{k + 1} } {\d x^{k + 1} } \ln x = \dfrac {k! \paren {-1}^k} {x^{k + 1} }$
Induction Step
This is our induction step:
\(\ds \frac {\d^{k + 1} } {\d x^{k + 1} } \ln x\) | \(=\) | \(\ds \map {\frac \d {\d x} } {\frac {\d^k} {\d x^k} \ln x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\frac \d {\d x} } {\paren {k - 1}! \paren {-1}^{k - 1} x^{-k} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k - 1}! \paren {-1}^{k - 1} \map {\frac \d {\d x} } {x^{-k} }\) | Derivative of Constant Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k - 1}! \paren {-1}^{k - 1} \paren {k x^{-k - 1} }\) | Power Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds k! \paren {-1}^k x^{-\paren {k + 1} }\) | Definition of Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k! \paren {-1}^k} {x^{k + 1} }\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$
$\blacksquare$