Math Doubts

Derivative Rule of Inverse Tan function

Formula

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

Introduction

The inverse tangent function is written as $\tan^{-1}{x}$ or $\arctan{(x)}$ in inverse trigonometry, where $x$ represents a real number. The derivative of the tan inverse function is written in mathematical form in differential calculus as follows.

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

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

The differentiation of the inverse tan function with respect to $x$ is equal to the reciprocal of the sum of one and $x$ squared.

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

Alternative forms

The differentiation of the tan inverse function can be written in terms of any variable. Here are some of the examples to learn how to express the formula for the derivative of inverse tangent function in calculus.

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

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

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

Proof

Learn how to derive the differentiation of the inverse tangent function from first principle.

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