Derivative of Reciprocal
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Theorem
Let $f: \R_{\ne 0} \to \R$ be the reciprocal function defined as:
- $\map f x = \dfrac 1 x$
Then its derivative is given by:
- $\map {f'} x = -\dfrac 1 {x^2}$
Proof
We have:
\(\ds \dfrac 1 x\) | \(=\) | \(\ds x^{-1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\dfrac \d {\d x} } {\dfrac 1 x}\) | \(=\) | \(\ds \paren {-1} x^{-2}\) | Power Rule for Derivatives | ||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 {x^2}\) |
$\blacksquare$
Sources
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Appendix $2$: Table of derivatives and integrals of common functions