Derivative of argsinh(x)

The derivative f’ of the function f(x) = argsinh(x) is given by f’(x) = 1 / sqrt(1 + x^2) for all x in R. To establish this outcome, we utilize the derivative of the inverse hyperbolic sine function. Let’s delve into the intricacies of this derivative.

Derivative of argsinh(x)

The derivative $f’$ of the function $f(x) = \operatorname{argsinh}(x)$ is expressed as: \(\forall x \in \mathbb{R}, \quad f'(x) = \dfrac{1}{\sqrt{1 + x^2}}\)

Proof

Firstly, recall that the function $\operatorname{argsinh}$ is the inverse function of $\sinh$:

\[\left(f^{-1} \circ f\right) = \left(\sinh \circ \operatorname{argsinh}\right)(x) = \sinh(\operatorname{argsinh}(x)) = x\]

By employing the result from the derivative of inverse functions, we have:

\[(g^{-1})'(x) = \dfrac{1}{g'(g^{-1}(x))}\]

Substituting $g^{-1} = f = \operatorname{argsinh}$ and $g = f^{-1} = \sinh$ into the formula above, we obtain:

\[f'(x) = \dfrac{1}{\sinh'(\operatorname{argsinh}(x))}\]

Given the derivative of hyperbolic sine: Derivative of sinh x, where $g(x) = \sinh x$, we have: \(\forall x \in \mathbb{R}, \quad g'(x) = \cosh x\)

Therefore, we deduce:

\[\begin{aligned} f'(x) &= \dfrac{1}{\sinh'(\operatorname{argsinh}(x))} \\ &= \dfrac{1}{\cosh(\operatorname{argsinh}(x))} \\ &= \dfrac{1}{\cosh(\operatorname{argsinh}(x))} \end{aligned}\]

Notably, $\cosh^2 x - \sinh^2 x = 1$, and by definition:

\[\left(f^{-1} \circ f\right) = \left(\sinh \circ \operatorname{argsinh} \right)(x) = {\color{red}{\sinh(\operatorname{argsinh}(x)) = x}}\]

Setting $x = \operatorname{argsinh}(x)$, we have:

\[\begin{aligned} 1 &= \cosh^2 x - {\color{red}{\sinh^2 x}} \\ &= \cosh^2(\operatorname{argsinh}(x)) - {\color{red}{x^2}} \end{aligned}\]

From this, we derive:

\[\cosh^2(\operatorname{argsinh}(x)) = 1 + x^2 \implies \cosh(\operatorname{argsinh}(x)) = \sqrt{1 + x^2}\]

The function $\operatorname{argsinh}(x)$ is defined for all $x \in \mathbb{R}$, and since $\operatorname{argsinh}(x)$ is unrestricted, the only valid solution is:

\[f'(x) = \dfrac{1}{\cosh(\operatorname{argsinh}(x))} = \dfrac{1}{\sqrt{1+x^2}}\]

Hence, the derivative of $\operatorname{argsinh}(x)$ is $\dfrac{1}{\sqrt{1+x^2}}$ for all $x \in \mathbb{R}$.