Derivative of Inverse Hyperbolic Sine

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Theorem

Let $x \in \R$ be a real number.

Let $\arsinh x$ denote the inverse hyperbolic sine of $x$.


Then:

$\map {\dfrac \d {\d x} } {\arsinh x} = \dfrac 1 {\sqrt {x^2 + 1} }$


Proof

\(\ds y\) \(=\) \(\ds \arsinh x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \sinh y\) Definition of Real Inverse Hyperbolic Sine
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d y}\) \(=\) \(\ds \cosh y\) Derivative of Hyperbolic Sine
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac 1 {\cosh y}\) Derivative of Inverse Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \pm \frac 1 {\sqrt {\sinh^2 y + 1} }\) Difference of Squares of Hyperbolic Cosine and Sine

For all $x \in \R$ we have that $\cosh y \ge 1$.

Thus it follows that it is necessary to take the positive root of $\sqrt {\sinh^2 y + 1}$.


So:

\(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac 1 {\sqrt {\sinh^2 y + 1} }\)
\(\ds \leadsto \ \ \) \(\ds \map {\frac \d {\d x} } {\arsinh x}\) \(=\) \(\ds \frac 1 {\sqrt {x^2 + 1} }\) Definition of $x$ and $y$

$\blacksquare$


Sources