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

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.

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved