Derivatives of Inverse Trigonometric Functions (College Board AP® Calculus BC): Study Guide

Derivatives of inverse trigonometric functions

How do I differentiate inverse trig functions?

• The inverse trigonometric functions (${\sin }^{-1}x$ etc) can be differentiated using:

  • The inverse function theorem, written as either

    • $\frac{dy}{dx}=\frac{1}{\left({\displaystyle \frac{dx}{dy}}\right)}$

    • or $g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}$ where $g\left(x\right)=f^{-1}\left(x\right)$

  • The chain rule

  • Trigonometric identities

• Note that you may also see inverse trig functions referred to with "arc" notation

  • E.g. ${\sin }^{-1}x=\arcsin x$

How do I differentiate inverse sine?

• Using the inverse function theorem, $g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}$ where $g\left(x\right)=f^{-1}\left(x\right)$

• Let $g\left(x\right)={\sin }^{-1}x$ and $f\left(x\right)=\sin x$

  • So $f^{\text{'}}\left(x\right)=\cos x$

• $g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}=\frac{1}{\cos \left({\sin }^{-1}x\right)}$

• Recall the identity ${\sin }^2\theta +{\cos }^2\theta \equiv 1$

  • This rearranges to $\cos \theta =\pm \sqrt{1-{\sin }^2\theta }$

• Substitute $\theta ={\sin }^{-1}x$

  • $\cos \left({\sin }^{-1}x\right)=\pm \sqrt{1-{\sin }^2\left({\sin }^{-1}x\right)}$

    • Note that $-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$ and $\cos x$ is non-negative on the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

    • Therefore, you can ignore the negative root

• Use this identity for the denominator

  • $g^{\text{'}}\left(x\right)=\frac{1}{\sqrt{1-{\sin }^2\left({\sin }^{-1}x\right)}}$

  • ${\sin }^2\left({\sin }^{-1}x\right)$ is the same as ${\left(\sin \left({\sin }^{-1}x\right)\right)}^2=x^2$

  • $g^{\text{'}}\left(x\right)=\frac{1}{\sqrt{1-x^2}}$

• $\frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$

• This result is only true for when $\sin x$ has an inverse

  • The domain for ${\sin }^{-1}x$ is $-1\le x\le 1$ or $\left|x\right|\le 1$

  • However the derivative of ${\sin }^{-1}x$ is only defined for $-1<x<1$ or $\left|x\right|<1$

    • This can be seen by inspecting the denominator of $g^{\text{'}}\left(x\right)$

    • The derivative becomes unbounded for $x=1$ or $x=-1$

How do I differentiate inverse cosine?

• Using the inverse function theorem, $g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}$ where $g\left(x\right)=f^{-1}\left(x\right)$

• Let $g\left(x\right)={\cos }^{-1}x$ and $f\left(x\right)=\cos x$

  • So $f^{\text{'}}\left(x\right)=-\sin x$

• $g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}=\frac{1}{-\sin \left({\cos }^{-1}x\right)}$

• Recall the identity ${\sin }^2\theta +{\cos }^2\theta \equiv 1$

  • This rearranges to $\sin \theta =\pm \sqrt{1-{\cos }^2\theta }$

• Substitute $\theta ={\cos }^{-1}x$

  • $\sin \left({\cos }^{-1}x\right)=\pm \sqrt{1-{\cos }^2\left({\cos }^{-1}x\right)}$

    • Note that $0\le {\cos }^{-1}x\le \pi$ and $\sin x$ is non-negative on the interval $\left[0,\pi \right]$

    • Therefore, you can ignore the negative root

• Use this identity for the denominator

  • $g^{\text{'}}\left(x\right)=\frac{1}{-\sqrt{1-{\cos }^2\left({\cos }^{-1}x\right)}}$

  • ${\cos }^2\left({\cos }^{-1}\left(x\right)\right)$ is the same as ${\left(\cos \left({\cos }^{-1}x\right)\right)}^2=x^2$

  • $g^{\text{'}}\left(x\right)=\frac{1}{-\sqrt{1-x^2}}$

• $\frac{d}{dx}\left({\cos }^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}}$

• This result is only true for when $\cos x$ has an inverse

  • The domain for ${\cos }^{-1}x$ is $-1\le x\le 1$ or $\left|x\right|\le 1$

  • However the derivative of ${\cos }^{-1}x$ is only defined for $-1<x<1$ or $\left|x\right|<1$

    • This can be seen by inspecting the denominator of $g^{\text{'}}\left(x\right)$

    • The derivative becomes unbounded for $x=1$ or $x=-1$

How do I differentiate inverse tangent?

• Using the inverse function theorem, $g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}$ where $g\left(x\right)=f^{-1}\left(x\right)$

• Let $g\left(x\right)={\tan }^{-1}x$ and $f\left(x\right)=\tan x$

  • So $f^{\text{'}}\left(x\right)={\sec }^{2}x$

• $g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}=\frac{1}{{\sec }^2\left({\tan }^{-1}x\right)}$

• Recall the identity ${\tan }^2\theta +1\equiv {\sec }^2\theta$

  • This can be derived by dividing the identity ${\sin }^2\theta +{\cos }^2\theta \equiv 1$ by ${\cos }^2\theta$

• Use this identity for the denominator

  • $g^{\text{'}}\left(x\right)=\frac{1}{{\tan }^2\left({\tan }^{-1}x\right)+1}$

  • ${\tan }^2\left({\tan }^{-1}\left(x\right)\right)$ is the same as ${\left(\tan \left({\tan }^{-1}x\right)\right)}^2=x^2$

  • $g^{\text{'}}\left(x\right)=\frac{1}{x^2+1}$

• $\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$

• This result is only true for when $\tan x$ has an inverse

  • The domain for ${\tan }^{-1}x$ is all real numbers

  • The derivative of $\tan x$ is also defined for all real numbers $x$

    • The derivative goes to zero as $x$ goes to $\pm \infty$

Summary of derivatives of inverse trig functions

• The methods above show how to find the derivatives of the three most common inverse trig functions

• The derivatives of the inverses of the reciprocal trig functions can be found in a similar way

• The table below summarizes the derivatives of all six inverse trig functions

Table of derivatives of inverse trig functions