# 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.

Latest Math Topics
Jun 26, 2023

###### Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Practice now

###### Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

###### Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.