# Derivatives of Inverse Trigonometric Functions

### Introduction to Inverse Trigonometric Functions

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

${\color{blue}{\sin x,\;}}\kern0pt\color{red}{\cos x,\;}\kern0pt\color{darkgreen}{\tan x,\;}\kern0pt\color{magenta}{\cot x,\;}\kern0pt\color{chocolate}{\sec x,\;}\kern0pt\color{maroon}{\csc x.\;}$

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

${\color{blue}{\arcsin x,\;}}\kern0pt \color{red}{\arccos x,\;}\kern0pt\color{darkgreen}{\arctan x,\;}\kern0pt\color{magenta}{\text{arccot }x,\;}\kern0pt\color{chocolate}{\text{arcsec }x,\;}\kern0pt\color{maroon}{\text{arccsc }x.\;}$

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 $$– \large{\frac{\pi }{2}}\normalsize$$ to $$\large{\frac{\pi }{2}}\normalsize$$ 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

${{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) = \frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} } = {\frac{1}{{\cos y}} } = {\frac{1}{{\sqrt {1 – {\sin^2}y} }} } = {\frac{1}{{\sqrt {1 – {\sin^2}\left( {\arcsin x} \right)} }} } = {\frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right).}$

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

${{\left( {\arccos x} \right)^\prime } }={ \frac{1}{{{{\left( {\cos y} \right)}^\prime }}} } = {\frac{1}{{\left( { – \sin y} \right)}} } = {- \frac{1}{{\sqrt {1 – {{\cos }^2}y} }} } = {- \frac{1}{{\sqrt {1 – {{\cos }^2}\left( {\arccos x} \right)} }} } = {- \frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right),}\qquad$

${{\left( {\arctan x} \right)^\prime } }={ \frac{1}{{{{\left( {\tan y} \right)}^\prime }}} } = {\frac{1}{{\frac{1}{{{{\cos }^2}y}}}} } = {\frac{1}{{1 + {{\tan }^2}y}} } = {\frac{1}{{1 + {{\tan }^2}\left( {\arctan x} \right)}} } = {\frac{1}{{1 + {x^2}}},}$

${\left( {\text{arccot }x} \right)^\prime } = {\frac{1}{{{{\left( {\cot y} \right)}^\prime }}}} = \frac{1}{{\left( { – \frac{1}{{{\sin^2}y}}} \right)}} = – \frac{1}{{1 + {{\cot }^2}y}} = – \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}} = – \frac{1}{{1 + {x^2}}},$

${{\left( {\text{arcsec }x} \right)^\prime } = {\frac{1}{{{{\left( {\sec y} \right)}^\prime }}} }} = {\frac{1}{{\tan y\sec y}} } = {\frac{1}{{\sec y\sqrt {{{\sec }^2}y – 1} }} } = {\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$

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 $$y \in \left( {0,{\large\frac{\pi }{2}\normalsize}} \right) \cup \left( {{\large\frac{\pi }{2}\normalsize},\pi } \right),$$ 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

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

$y = \arcsin \left( {x – 1} \right)$

### Example 3

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

### Example 4

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

### Example 5

${y = \arctan \frac{{x + 1}}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ne 1} \right)}$

### Example 6

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

### Example 7

$y = \arctan \left( {x – \sqrt {1 + {x^2}} } \right)$

### Example 8

Calculate the derivative of the function $$y = \arccos x\arctan x$$ at $$x = 0.$$

### Example 9

Using the chain rule, derive the formula for the derivative of the inverse sine function.

### Example 10

$y = \arcsin \sqrt {1 – {x^2}}$

### Example 11

Calculate the derivative of $$y = \arctan \sqrt {{e^x}}$$ at $$x = 0.$$

### Example 12

Compute the derivative of the function $$y = \arcsin \large{\frac{1}{{\sqrt x }}}\normalsize$$ at $$x = 2.$$

### Example 13

$y = \frac{2}{{\sqrt 3 }}\text{arccot}\frac{{2x + 1}}{{\sqrt 3 }}$

### Example 14

Show that $${\large\frac{d}{{dx}}\normalsize} \left( {x\arcsin x + \sqrt {1 – {x^2}} } \right)$$ $$= \arcsin x.$$

### Example 15

$y = \arcsin \frac{{1 – {x^2}}}{{1 + {x^2}}}$

### Example 16

$y = \frac{1}{x}\text{arccsc}\frac{1}{x}$

### Example 17

$y = \arccos \left( {\cos x} \right)$

### Example 18

Determine the derivative of the function $$y = \arcsin \large{\frac{{2x}}{{1 + {x^2}}}}\normalsize$$ at $$x = 2.$$

### 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.

${y = \arctan \frac{{x + 1}}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ne 1} \right)}$

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.

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

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Problems 1-6
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Problems 7-18