$\dfrac{d}{dx}{\, \sinh^{-1}{x}}$ $\,=\,$ $\dfrac{1}{\sqrt{x^2+1}}$

The inverse hyperbolic sine function is written as $\sinh^{-1}{(x)}$ or $\operatorname{arcsinh}{(x)}$ in mathematics when the $x$ represents a variable. The derivative of the inverse hyperbolic sine function with respect to $x$ is written in the following mathematical forms.

$(1).\,\,\,$ $\dfrac{d}{dx}{\, (\sinh^{-1}{x})}$

$(2).\,\,\,$ $\dfrac{d}{dx}{\, (\operatorname{arcsinh}{x})}$

Mathematically, the derivative of the inverse hyperbolic sine function is simply written as $(\sinh^{-1}{x})’$ or $(\operatorname{arcsinh}{x})’$ in differential calculus.

The differentiation of the hyperbolic inverse sin function with respect to $x$ is equal to multiplicative inverse of square root of sum of $1$ and $x$ squared.

$\implies$ $\dfrac{d}{dx}{\, \sinh^{-1}{x}}$ $\,=\,$ $\dfrac{1}{\sqrt{x^2+1}}$

In mathematics, the derivative of inverse hyperbolic sine function can also be written in terms of any variable.

$(1) \,\,\,$ $\dfrac{d}{db}{\, \sinh^{-1}{b}}$ $\,=\,$ $\dfrac{1}{\sqrt{b^2+1}}$

$(2) \,\,\,$ $\dfrac{d}{dl}{\, \sinh^{-1}{l}}$ $\,=\,$ $\dfrac{1}{\sqrt{l^2+1}}$

$(3) \,\,\,$ $\dfrac{d}{dy}{\, \sinh^{-1}{y}}$ $\,=\,$ $\dfrac{1}{\sqrt{y^2+1}}$

Learn how to prove the differentiation formula of hyperbolic inverse sine function by the first principle of differentiation.

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