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.
