# Derivatives of Inverse Trigonometric Functions

## Introduction to Inverse Trigonometric Functions

In the previous topic, we have learned the derivatives of six basic trigonometric functions:

$\color{blue}{\sin x,\;} \color{red}{\cos x,\;} \color{darkgreen}{\tan x,\;} \color{magenta}{\cot x,\;} \color{chocolate}{\sec x,\;} \color{maroon}{\csc x.\;}$

In this section, we are going to look at the derivatives of the inverse trigonometric functions, which are respectively denoted as

$\color{blue}{\arcsin x,\;} \color{red}{\arccos x,\;} \color{darkgreen}{\arctan x,\;} \color{magenta}{\text{arccot }x,\;} \color{chocolate}{\text{arcsec }x,\;} \color{maroon}{\text{arccsc }x.\;}$

The inverse functions exist when appropriate restrictions are placed on the domain of the original functions.

For example, the domain for $$\arcsin x$$ is from $$-1$$ to $$1.$$ The range, or output for $$\arcsin x$$ is all angles from $$- \frac{\pi }{2}$$ to $$\frac{\pi }{2}$$ radians.

The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined.

## Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. For example, the sine function $$x = \varphi \left( y \right)$$ $$= \sin y$$ is the inverse function for $$y = f\left( x \right)$$ $$= \arcsin x.$$ Then the derivative of $$y = \arcsin x$$ is given by

$\left( {\arcsin x} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\sin y} \right)}^\prime }}} = \frac{1}{{\cos y}} = \frac{1}{{\sqrt {1 - {\sin^2}y} }} = \frac{1}{{\sqrt {1 - {\sin^2}\left( {\arcsin x} \right)} }} = \frac{1}{{\sqrt {1 - {x^2}} }}\;\;\left( { - 1 \lt x \lt 1} \right).$

Using this technique, we can find the derivatives of the other inverse trigonometric functions:

$\left( {\arccos x} \right)^\prime = \frac{1}{{{{\left( {\cos y} \right)}^\prime }}} = \frac{1}{{\left( { - \sin y} \right)}} = - \frac{1}{{\sqrt {1 - {{\cos }^2}y} }} = - \frac{1}{{\sqrt {1 - {{\cos }^2}\left( {\arccos x} \right)} }} = - \frac{1}{{\sqrt {1 - {x^2}} }}\;\;\left( { - 1 \lt x \lt 1} \right),$
$\left( {\arctan x} \right)^\prime = \frac{1}{{{{\left( {\tan y} \right)}^\prime }}} = \frac{1}{{\frac{1}{{{{\cos }^2}y}}}} = \frac{1}{{1 + {{\tan }^2}y}} = \frac{1}{{1 + {{\tan }^2}\left( {\arctan x} \right)}} = \frac{1}{{1 + {x^2}}},$
$\left( {\text{arccot }x} \right)^\prime = \frac{1}{{{{\left( {\cot y} \right)}^\prime }}} = \frac{1}{{\left( { - \frac{1}{{{\sin^2}y}}} \right)}} = - \frac{1}{{1 + {{\cot }^2}y}} = - \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}} = - \frac{1}{{1 + {x^2}}},$
${\left( {\text{arcsec }x} \right)^\prime = \frac{1}{{{{\left( {\sec y} \right)}^\prime }}} } = \frac{1}{{\tan y\sec y}} = \frac{1}{{\sec y\sqrt {{{\sec }^2}y - 1} }} = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}.$

In the last formula, the absolute value $$\left| x \right|$$ in the denominator appears due to the fact that the product $${\tan y\sec y}$$ should always be positive in the range of admissible values of $$y$$, where $$y \in \left( {0,{\frac{\pi }{2}}} \right) \cup \left( {{\frac{\pi }{2}},\pi } \right),$$ that is the derivative of the inverse secant is always positive.

Similarly, we can obtain an expression for the derivative of the inverse cosecant function:

$\left( {\text{arccsc }x} \right)^\prime = \frac{1}{{{{\left( {\csc y} \right)}^\prime }}} = -\frac{1}{{\cot y\csc y}} = -\frac{1}{{\csc y\sqrt {{{\csc }^2}y - 1} }} = -\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}.$

## Table of Derivatives of Inverse Trigonometric Functions

The derivatives of $$6$$ inverse trigonometric functions considered above are consolidated in the following table:

In the examples below, find the derivative of the given function.

## Solved Problems

### Example 1.

$y = \arctan {\frac{1}{x}}$

Solution.

By the chain rule,

$y^\prime = \left( {\arctan \frac{1}{x}} \right)^\prime = \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \cdot \left( {\frac{1}{x}} \right)^\prime = \frac{1}{{1 + \frac{1}{{{x^2}}}}} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = - \frac{{{x^2}}}{{\left( {{x^2} + 1} \right){x^2}}} = - \frac{1}{{1 + {x^2}}}.$

### Example 2.

$y = \arcsin \left( {x - 1} \right)$

Solution.

$\require{cancel} y^\prime = \left( {\arcsin \left( {x - 1} \right)} \right)^\prime = \frac{1}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }} = \frac{1}{{\sqrt {1 - \left( {{x^2} - 2x + 1} \right)} }} = \frac{1}{{\sqrt {\cancel{1} - {x^2} + 2x - \cancel{1}} }} = \frac{1}{{\sqrt {2x - {x^2}} }}.$

### Example 3.

$y = \text{arccot}\,{x^2}$

Solution.

Using the chain rule, we have

$y^\prime = \left( {\text{arccot}\,{x^2}} \right)^\prime = - \frac{1}{{1 + {{\left( {{x^2}} \right)}^2}}} \cdot \left( {{x^2}} \right)^\prime = - \frac{{2x}}{{1 + {x^4}}}.$

### Example 4.

$y ={\frac{1}{a}} \arctan {\frac{x}{a}}$

Solution.

By the chain rule,

$y^\prime = \left( {\frac{1}{a}\arctan \frac{x}{a}} \right)^\prime = \frac{1}{a} \cdot \frac{1}{{1 + {{\left( {\frac{x}{a}} \right)}^2}}} \cdot \left( {\frac{x}{a}} \right)^\prime = \frac{1}{a} \cdot \frac{1}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \cdot \frac{1}{a} = \frac{1}{{{a^2}}} \cdot \frac{{{a^2}}}{{{a^2} + {x^2}}} = \frac{1}{{{a^2} + {x^2}}}.$

### Example 5.

$y = \arctan \frac{{x + 1}}{{x - 1}}\;\left( {x \ne 1} \right)$

Solution.

By the chain and quotient rules, we have

$y'\left( x \right) = \left( {\arctan \frac{{x + 1}}{{x - 1}}} \right)^\prime = \frac{1}{{1 + {{\left( {\frac{{x + 1}}{{x - 1}}} \right)}^2}} \cdot \left( {\frac{{x + 1}}{{x - 1}}} \right)^\prime } = \frac{{1 \cdot \left( {x - 1} \right) - \left( {x + 1} \right) \cdot 1}}{{{{\left( {x - 1} \right)}^2} + {{\left( {x + 1} \right)}^2}}} = \frac{{\cancel{\color{blue}{x}} - \color{red}{1} - \cancel{\color{blue}{x}} - \color{red}{1}}}{{\color{maroon}{x^2} - \cancel{\color{green}{2x}} + \color{DarkViolet}{1} + \color{maroon}{x^2} + \cancel{\color{green}{2x}} + \color{DarkViolet}{1}}} = \frac{{ - \color{red}{2}}}{{\color{maroon}{2{x^2}} + \color{DarkViolet}{2}}} = - \frac{1}{{1 + {x^2}}}.$

### Example 6.

$y = \text{arccot}{\frac{1}{{{x^2}}}}$

Solution.

$y^\prime = \left( {\text{arccot}\frac{1}{{{x^2}}}} \right)^\prime = - \frac{1}{{1 + {{\left( {\frac{1}{{{x^2}}}} \right)}^2}}} \cdot \left( {\frac{1}{{{x^2}}}} \right)^\prime = - \frac{1}{{1 + \frac{1}{{{x^4}}}}} \cdot \left( { - 2{x^{ - 3}}} \right) = \frac{{2{x^4}}}{{\left( {{x^4} + 1} \right){x^3}}} = \frac{{2x}}{{1 + {x^4}}}.$

See more problems on Page 2.