Math Doubts

Derivative Rule of Inverse Cosine function

Formula

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

Introduction

When $x$ represents a variable, the inverse cosine function is written as $\cos^{-1}{(x)}$ or $\arccos{(x)}$ in inverse trigonometry. In differential calculus, the derivative of the cos inverse function with respect to $x$ is written in following two mathematical forms.

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

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

The derivative of the inverse cos function with respect to $x$ is equal to the negative reciprocal of the square root of the subtraction of square of $x$ from one.

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

Alternative forms

The differentiation of the cos inverse function can be written in any variable. Here are few examples to learn how to write the formula for the derivative of cosine inverse function in differential calculus.

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

$(2) \,\,\,$ $\dfrac{d}{du}{\,\Big(\cos^{-1}{(u)}\Big)}$ $\,=\,$ $-\dfrac{1}{\sqrt{1-u^2}}$

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

Proof

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

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