Math Doubts

Derivative Rule of Inverse Sine function

Formula

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

Introduction

Let $x$ be a variable. The inverse sine function is written as $\sin^{-1}{(x)}$ or $\arcsin{(x)}$ in inverse trigonometry. In differential calculus, the differentiation of the sin inverse function is written in mathematical form as follows.

$(1) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\sin^{-1}{(x)}\Big)}$

$(2) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\arcsin{(x)}\Big)}$

The differentiation of the inverse sin function with respect to $x$ is equal to the reciprocal of the square root of the subtraction of square of $x$ from one.

$\implies$ $\dfrac{d}{dx}{\, \Big(\sin^{-1}{(x)}\Big)}$ $\,=\,$ $\dfrac{1}{\sqrt{1-x^2}}$

Alternative forms

The derivative of the sin inverse function can be written in terms of any variable. Here, some of the examples are given to learn how to express the formula for the derivative of inverse sine function in differential calculus.

$(1) \,\,\,$ $\dfrac{d}{dy}{\, \Big(\sin^{-1}{(y)}\Big)}$ $\,=\,$ $\dfrac{1}{\sqrt{1-y^2}}$

$(2) \,\,\,$ $\dfrac{d}{dm}{\, \Big(\sin^{-1}{(m)}\Big)}$ $\,=\,$ $\dfrac{1}{\sqrt{1-m^2}}$

$(3) \,\,\,$ $\dfrac{d}{dz}{\, \Big(\sin^{-1}{(z)}\Big)}$ $\,=\,$ $\dfrac{1}{\sqrt{1-z^2}}$

Proof

Learn how to derive the derivative of the inverse sine function formula by first principle.

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