Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.7: Inverse Trigonometric Derivatives

  • Page ID
    529
  •  

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    Definition of the Inverse Trig Functions

    Recall that we write \( \sin^{-1} x\) or \(\text{arcsin}\, x\) to mean the inverse \(\sin\) of \(x\) restricted to have values between \(-\pi/2\) and \(\pi/2\) (Note that \(\sin x\) does not pass the horizontal line test, hence we need to restrict the domain.) We define the other five inverse trigonometric functions similarly.

    Inverse of Arctrig Functions

    Example 1

    Find \(\tan(\sin^{-1} x)\)

    Solution

    \[\tan(\sin^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.\]

    right triangle:  Hyp = 1, Opp = x, Adj = root(1-x^2)

    The triangle above demonstrates that

    \[\sin t = \dfrac{x}{1} = \dfrac{opp}{hyp}.\]

    Hence 

    \[ \tan(\tan^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.\]

    Since the

    \[ \text{tangent} = \dfrac{opp}{adj}.\]

    We have

    \[ \tan ( \sin^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.\]

    Exercise

    Simplify

    \[ \cos(\tan^{-1} (2x)).\]

    Derivatives of the Arctrigonometric Functions

    Recall that if \(f\) and \(g\) are inverses, then

    \[ g'(x) \dfrac{1}{f'(g(x))}.\]

    What is 

    \[ \dfrac{d}{dx} \tan^{-1} x\text{?}\]

    We use the formula:

    \[\frac{d}{dx} \tan^{-1} x= \dfrac{1}{\sec^2 (\tan^{-1} x)} = \cos^2 (\tan^{-1} x).\]

    Since

    \[ \tan q = \dfrac{opp}{adj} = \dfrac{x}{1} \]

    we have 

    \[ hyp = \sqrt{1+x^2}\]

    so that

    \[ \cos^2 (\tan^{-1} x) = \left( \dfrac{1}{\sqrt{1+x^2}}\right)^2 = \dfrac{1}{1+x^2}.\]

    Relationships

    \[ \dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2} \]

    \[ \dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}} \]

    \[ \dfrac{d}{dx} \sec^{-1} x = \dfrac{1}{|x|\sqrt{x^2-1|}} \]

    Recall that

    \[ \cos x = \sin \left(\dfrac{\pi}{2} - x \right) \]

    hence

    \[ \cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x\]

    so

    \[ \dfrac{d}{dx} \cos^{-1} x = \dfrac{d}{dx} \left[ \dfrac{\pi}{2} - \sin^{-1} x \right] \]

    \[ = \dfrac{-d}{dx} \sin^{-1} x = \dfrac{-1}{\sqrt{1-x^2}}.\]

    Similarly:

    \[\dfrac{d}{dx} \cot^{-1} x = \dfrac{-1}{\sqrt{1 + x^2}} \]

    \[\dfrac{d}{dx} \text{csc}\, x\ = \dfrac{-1}{\sqrt{1-x^2}}. \]

    Example 2

    Find the derivative of \( \cos(\sin^{-1} x)\).

    Solution

    Let \(y = \cos u\) , \(u = \sin^{-1} x\), and \(y' = -\sin u\)

    \[ y'u= -\sin (\sin^{-1} x) = x\]

    \[ u' = \dfrac{1}{\sqrt{1-x^2}}.\]

    We arrive at

    \[ \dfrac{dy}{dx} = \dfrac{x}{\sqrt{1-x^2}}.\]

    Contributors