Calculus

Differentiation of Functions

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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:

    derivatives of inverse trigonometric functions

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


  • Solved Problems

    Click 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