4.7: Inverse Trigonometric Derivatives
- Page ID
- 529
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
Find \(\tan(\sin^{-1} x)\)
Solution
\[\tan(\sin^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.\nonumber \]
The triangle above demonstrates that
\[\sin t = \dfrac{x}{1} = \dfrac{opp}{hyp}.\nonumber \]
Hence
\[ \tan(\tan^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.\nonumber \]
Since the
\[ \text{tangent} = \dfrac{opp}{adj}.\nonumber \]
We have
\[ \tan ( \sin^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.\nonumber \]
Simplify
\[ \cos(\tan^{-1} (2x)).\nonumber \]
Derivatives of the Arctrigonometric Functions
Recall that if \(f\) and \(g\) are inverses, then
\[ g'(x) \dfrac{1}{f'(g(x))}.\nonumber \]
What is
\[ \dfrac{d}{dx} \tan^{-1} x\text{?}\nonumber \]
We use the formula:
\[\frac{d}{dx} \tan^{-1} x= \dfrac{1}{\sec^2 (\tan^{-1} x)} = \cos^2 (\tan^{-1} x).\nonumber \]
Since
\[ \tan q = \dfrac{opp}{adj} = \dfrac{x}{1} \nonumber \]
we have
\[ hyp = \sqrt{1+x^2}\nonumber \]
so that
\[ \cos^2 (\tan^{-1} x) = \left( \dfrac{1}{\sqrt{1+x^2}}\right)^2 = \dfrac{1}{1+x^2}.\nonumber \]
Relationships
\[ \dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2} \nonumber \]
\[ \dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}} \nonumber \]
\[ \dfrac{d}{dx} \sec^{-1} x = \dfrac{1}{|x|\sqrt{x^2-1|}} \nonumber \]
Recall that
\[ \cos x = \sin \left(\dfrac{\pi}{2} - x \right) \nonumber \]
hence
\[ \cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x\nonumber \]
so
\[ \dfrac{d}{dx} \cos^{-1} x = \dfrac{d}{dx} \left[ \dfrac{\pi}{2} - \sin^{-1} x \right] \nonumber \]
\[ = \dfrac{-d}{dx} \sin^{-1} x = \dfrac{-1}{\sqrt{1-x^2}}.\nonumber \]
Similarly:
\[\dfrac{d}{dx} \cot^{-1} x = \dfrac{-1}{\sqrt{1 + x^2}} \nonumber \]
\[\dfrac{d}{dx} \text{csc}\, x\ = \dfrac{-1}{\sqrt{1-x^2}}. \nonumber \]
Find the derivative of \( \cos(\sin^{-1} x)\).
Solution
Let \(y = \cos u\), \(u = \sin^{-1} x\), and \(y' = -\sin u\)
\[ y'u= -\sin (\sin^{-1} x) = x\nonumber \]
\[ u' = \dfrac{1}{\sqrt{1-x^2}}.\nonumber \]
We arrive at
\[ \dfrac{dy}{dx} = \dfrac{x}{\sqrt{1-x^2}}.\nonumber \]
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.