Derivative of Inverse Hyperbolic Cotangent Function

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Theorem

Let $u$ be a differentiable real function of $x$.

Then:

$\map {\dfrac \d {\d x} } {\coth^{-1} u} = \dfrac {-1} {u^2 - 1} \dfrac {\d u} {\d x}$

where $\size u > 1$

where $\coth^{-1}$ is the inverse hyperbolic cotangent.


Proof

\(\ds \map {\frac \d {\d x} } {\coth^{-1} u}\) \(=\) \(\ds \map {\frac \d {\d u} } {\coth^{-1} u} \frac {\d u} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac {-1} {u^2 - 1} \frac {\d u} {\d x}\) Derivative of Inverse Hyperbolic Cotangent

$\blacksquare$


Also presented as

Can also be seen (and in fact most often seen) presented in the form:

$\map {\dfrac \d {\d x} } {\coth^{-1} u} = \dfrac 1 {1 - u^2} \dfrac {\d u} {\d x}$

but this obscures the fact that $\size u > 1$.

The condition is also often seen presented as $u^2 > 1$


Also see


Sources

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