2.9 Derivatives of Inverse Trig Functions

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\newrobustcmd }[0]{} \newcommand {\csshow }[1]{\begingroup \expandafter \endgroup \expandafter \show \csname #1\endcsname } \newcommand {\csmeaning }[1]{\ifcsname #1\endcsname \expandafter \meaning \csname #1\endcsname \else \detokenize {undefined}\fi } \newcommand {\ifdefmacro }[0]{} \newcommand {\ifdefparam }[0]{} \newcommand {\ifdefprotected }[0]{} \newcommand {\ifnumequal }[1]{\ifnumcomp {#1}=} \newcommand {\ifnumgreater }[1]{\ifnumcomp {#1}>} \newcommand {\ifnumless }[1]{\ifnumcomp {#1}<} \newcommand {\ifdimequal }[1]{\ifdimcomp {#1}=} \newcommand {\ifdimgreater }[1]{\ifdimcomp {#1}>} \newcommand {\ifdimless }[1]{\ifdimcomp {#1}<} \newcommand {\expandonce }[1]{\unexpanded \expandafter {#1}} \newcommand {\csexpandonce }[1]{\expandafter \expandonce \csname #1\endcsname } \newcommand {\protecting }[0]{} \newcommand {\csdef }[1]{\expandafter \def \csname #1\endcsname } \newcommand {\csedef }[1]{\expandafter \edef \csname #1\endcsname } \newcommand {\csgdef }[1]{\expandafter \gdef \csname #1\endcsname } \newcommand {\csxdef }[1]{\expandafter \xdef \csname #1\endcsname } \newcommand {\cslet }[2]{\expandafter \let \csname #1\endcsname #2} \newcommand {\letcs }[2]{\ifcsdef {#2} {\expandafter \let \expandafter #1\csname #2\endcsname } {\undef #1}} \newcommand {\csletcs }[2]{\ifcsdef {#2} {\expandafter \let \csname #1\expandafter \endcsname \csname #2\endcsname } {\csundef {#1}}} \newcommand {\csuse }[1]{\ifcsname #1\endcsname \csname #1\expandafter \endcsname \fi } \newcommand {\appto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\eappto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{\expandonce #1#2}}} \newcommand {\gappto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\xappto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{\expandonce #1#2}}} \newcommand {\preto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\epreto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{#2\expandonce #1}}} \newcommand {\gpreto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\xpreto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{#2\expandonce #1}}} \newcommand {\csappto }[1]{\expandafter \appto \csname #1\endcsname } \newcommand {\cseappto }[1]{\expandafter \eappto \csname #1\endcsname } \newcommand {\csgappto }[1]{\expandafter \gappto \csname #1\endcsname } \newcommand {\csxappto }[1]{\expandafter \xappto \csname #1\endcsname } \newcommand {\cspreto }[1]{\expandafter \preto \csname #1\endcsname } \newcommand {\csepreto }[1]{\expandafter \epreto \csname #1\endcsname } \newcommand {\csgpreto }[1]{\expandafter \gpreto \csname #1\endcsname } \newcommand {\csxpreto }[1]{\expandafter \xpreto \csname #1\endcsname } \newcommand {\csnumdef }[1]{\expandafter \numdef \csname #1\endcsname } \newcommand {\csnumgdef }[1]{\expandafter \numgdef \csname #1\endcsname } \newcommand {\csdimdef }[1]{\expandafter \dimdef \csname #1\endcsname } \newcommand {\csdimgdef }[1]{\expandafter \dimgdef \csname #1\endcsname } \newcommand {\csgluedef }[1]{\expandafter \gluedef \csname #1\endcsname } \newcommand {\csgluegdef }[1]{\expandafter \gluegdef \csname #1\endcsname } \newcommand {\mudef }[2]{\ifundef #1{\def #1{0mu}}{}\edef #1{\the \muexpr #2}} \newcommand {\csmudef }[1]{\expandafter \mudef \csname #1\endcsname } \newcommand {\csmugdef }[1]{\expandafter \mugdef \csname #1\endcsname } \newcommand {\listbreak }[0]{} \newcommand {\listadd }[2]{\ifblank {#2}{}{\appto #1{#2|}}} \newcommand {\listgadd }[2]{\ifblank {#2}{}{\gappto #1{#2|}}} \newcommand {\listcsadd }[1]{\expandafter \listadd \csname #1\endcsname } \newcommand {\listcseadd }[1]{\expandafter \listeadd \csname #1\endcsname } \newcommand {\listcsgadd }[1]{\expandafter \listgadd \csname #1\endcsname } \newcommand {\listcsxadd }[1]{\expandafter \listxadd \csname #1\endcsname } \newcommand {\dolistloop }[0]{\forlistloop \do } \newcommand {\dolistcsloop }[0]{\forlistcsloop \do } \newcommand {\AfterPreamble }[0]{\AtBeginDocument } \newcommand {\tcolorboxenvironment }[2]{\BeforeBeginEnvironment {#1}{\begin {tcolorbox}[savedelimiter={#1},#2]}\AfterEndEnvironment {#1}{\end {tcolorbox}}} \newcommand {\tcbpkgprefix }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

In this section we compute derivatives involving $\tan ^{-1}(x), \sin ^{-1}(x)$ and $\cos ^{-1}(x)$ .

We begin by computing the derivative of the inverse trigonometric function $f(x) = \tan ^{-1}(x)$ . The following Pythagorean trigonometric identity will be needed:

$1 + \tan ^2(\theta ) = \sec ^2(\theta ).$ This identity follows from $\cos ^2(\theta ) + \sin ^2(\theta ) = 1$ by dividing both sides by $\cos ^2(\theta )$ .

We begin the derivation by using the fact that $\tan (x)$ and $\tan ^{-1}(x)$ are inverse functions, so that:

$\tan (\tan ^{-1}(x)) = x.$

We differentiate both sides of this equation with respect to $x$ :

$\dd {x} \tan (\tan ^{-1}(x)) = \ddx x.$

Using the Chain Rule on the left side gives:

$\sec ^2(\tan ^{-1}(x)) \ddx \left [\tan ^{-1}(x)\right ] = 1.$

Now, we can solve this for the derivative of $\tan ^{-1}(x)$ :

$\ddx \left [\tan ^{-1}(x)\right ] = \frac {1}{\sec ^2\left (\tan ^{-1}(x)\right )}.$

Next, we use the Pythagorean Identity:

$\ddx \left [\tan ^{-1}(x)\right ] = \frac {1}{1 + \tan ^2\left (\tan ^{-1}(x)\right )}.$

Finally, using the property of inverse functions:

$\ddx \left [\tan ^{-1}(x)\right ] = \frac {1}{1 + x^2}.$

Similar arguments can show that:

$\ddx \left [\sin ^{-1}(x)\right ] = \frac {1}{\sqrt {1 - x^2}} \;\;\text { and } \;\;\ddx \left [\cos ^{-1}(x)\right ] = -\frac {1}{\sqrt {1 - x^2}}.$

Below is a graph of $f(x) = \tan ^{-1}(x)$ (in blue) and its derivative, $f'(x) = \tfrac {1}{1+x^2}$ (in purple). Notice that the slope of the tangent line to $f(x)$ (in red) is the height of the corresponding point on $f'(x)$ . Use it to see a graphical representation of the answers to the problem above.

example 1

Find the equation of the tangent line to the graph of $f(x) = \tan ^{-1}(x)$ at $x=0.$
The point of tangency is $(0, f(0)) = (0, 0)$ since $\tan ^{-1}(0) = 0$ . The slope of the tangent line is given by $m = f'(0) = \tfrac {1}{1+0^2} = 1$ . To create the equation of the line we use the Point slope form: $y-y_0 = m(x-x_0)$ and we get $y-0 = 1(x-0)$ or $y=x$ .

(problem 1a) Find the equation of the tangent line to the graph of $f(x) = \tan ^{-1}(x)$ at $x=1.$

The point of tangency is $(1, f(1))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {x/2 + \pi /4 - 1/2}$

(problem 1b)

Find the equation of the tangent line to the graph of $f(x) = \sin ^{-1}(x)$ at $x=0.$

The point of tangency is $(0, f(0))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {x}$

(problem 1c)

Find the equation of the tangent line to the graph of $f(x) = \cos ^{-1}(x)$ at $x=0.$

The point of tangency is $(0, f(0))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {-x + \pi /2}$

example 2 Find $h'(x)$ if $h(x) = (1+x^2)\tan ^{-1}(x)$ .
We write $h(x)$ as a product: $h(x) = f(x)\cdot g(x)$ with $f(x) = 1+x^2 \quad \text {and} \quad g(x) = \tan ^{-1}(x).$ To use the product rule we need the derivatives: $f'(x) = 2x \quad \text {and} \quad g'(x) = \frac {1}{1+x^2}.$ We can now write the derivative: $\begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= 2x\tan^{-1}(x) + (1+x^2)\frac{1}{1+x^2} \\ &= 2x\tan^{-1}(x) + 1. \end{align*}$

Here is a video of the preceding example

(problem 2a) Compute $\frac {d}{dx} x^3 \tan ^{-1}(x)$

Use the Product Rule with $x^3$ and $\tan ^{-1}(x)$ .

$(fg)' = f'g+g'f$ .

The derivative of $x^3$ is $3x^2$

The derivative of $\tan ^{-1}(x)$ is $\frac {1}{1+x^2}$

The derivative of $x^3\tan ^{-1}(x)$ with respect to $x$ is $\answer {3x^2 \tan ^{-1}(x) + \frac {x^3}{1+x^2}}$

(problem 2b) Compute $\frac {d}{dx} (1-x^2) \sin ^{-1}(x)$

Use the Product Rule with $1-x^2$ and $\sin ^{-1}(x)$ .

$(fg)' = f'g+g'f$ .

The derivative of $1 - x^2$ is $-2x$

The derivative of $\sin ^{-1}(x)$ is $\frac {1}{\sqrt {1-x^2}}$

$(1-x^2) \frac {1}{\sqrt {1-x^2}} = \sqrt {1-x^2}$

The derivative of $(1-x^2)\sin ^{-1}(x)$ with respect to $x$ is $\answer {-2x \sin ^{-1}(x) + \sqrt {1-x^2}}$

The Chain Rule versions of these formulas are:

$\ddx \left [\tan ^{-1}(g(x))\right ] = \frac {g'(x)}{1 + g^2(x)},$ $\ddx \left [\sin ^{-1}(g(x))\right ] = \frac {g'(x)}{\sqrt {1 - g^2(x)}}$

and

$\ddx \left [\cos ^{-1}(g(x))\right ] = -\frac {g'(x)}{\sqrt {1 - g^2(x)}}.$

example 3 Find $h'(x)$ if $h(x) = \tan ^{-1}(4x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 4x \quad \text {and} \quad f(x) = \tan ^{-1}(x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 4; \quad \text {next}$ $f'(x) = \frac {1}{1+x^2} , \quad \text {so}$ $f'(g(x)) = \frac {1}{1+g^2(x)} =\frac {1}{1+(4x)^2}=\frac {1}{1+16x^2}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \frac {1}{1+16x^2} \cdot 4 = \frac {4}{1+16x^2}.$

Here is a video of the example

(problem 3a) Compute $\frac {d}{dx} \tan ^{-1}(3x)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \tan ^{-1}(x)$ and the inside function is $g(x) = 3x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\tan ^{-1}(3x)$ with respect to $x$ is $\answer {\frac {3}{1+(3x)^2}}$

(problem 3b) Find the equation of the tangent line to the graph of $f(x) = \cos (x)$ at $x=\pi /2.$

The point of tangency is $(\pi /2, f(\pi /2))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {-x + \pi /2}$