Derivative of Logarithm over Power

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

\(\ds \dfrac \d {\d x} \dfrac {\ln x} {x^n}\) \(=\) \(\ds \dfrac \d {\d x} x^{-n} \ln x\)
\(\ds \) \(=\) \(\ds \ln x \dfrac \d {\d x} x^{-n} + x^{-n} \dfrac \d {\d x} \ln x\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds - n x^{-n - 1} \ln x + x^{-n} \dfrac 1 x\) Primitive of Reciprocal, Primitive of Power
\(\ds \) \(=\) \(\ds -n x^{-n - 1} \ln x + x^{-n - 1}\)
\(\ds \) \(=\) \(\ds \frac {1 - n \ln x} {x^{n + 1} }\)

$\blacksquare$