Derivatives of inverse trigonometric functions

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \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]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \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 {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\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 {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We derive the derivatives of inverse trigonometric functions using implicit differentiation.

Now we will derive the derivative of arcsine, arctangent, and arcsecant.

The derivative of arcsine $\ddx \arcsin (x) = \frac {1}{\sqrt {1-x^2}}.$

Recall $\arcsin (x) = \theta$ means that $\sin (\theta ) = x$ and $\answer [given]{\frac {-\pi }{2}}\le \theta \le \answer [given]{\frac {\pi }{2}}$ . Implicitly differentiating with respect $x$ we see $\begin{align*} \sin(\theta) &= x\\ \ddx \sin(\theta) &= \ddx x &\text{Differentiate both sides.}\\ \cos(\theta) \cdot \theta' &= 1 &\text{Implicit differentiation.}\\ \theta' &= \frac{1}{\cos(\theta)} &\text{Solve for $\theta'$.} \end{align*}$

While $\theta ' = \frac {1}{\cos (\theta )}$ , we need our answer written in terms of $x$ . Since we are assuming that $\sin (\theta ) = x,$ consider the following triangle with the unit circle:

From the unit circle above, we see that $\begin{align*} \theta' &= \frac{1}{\cos(\theta)}\\ &=\frac{\mathrm{hyp}}{\mathrm{adj}}\\ &= \answer[given]{\frac{1}{\sqrt{1-x^2}}}. \end{align*}$

Recall, $\cos {x}\ge 0$ on the interval $\Bigl (-\frac {\pi }{2},\frac {\pi }{2}\Bigr )$ .

To be completely explicit, $\ddx \theta = \ddx \arcsin (x) = \answer [given]{\frac {1}{\sqrt {1-x^2}}}.$

Compute: $\ddx \sin ^{-1}(x) \begin {prompt} = \answer [given]{1/\sqrt {1-x^2}} \end {prompt}$

We can do something similar with arctangent.

The derivative of arctangent $\ddx \arctan (x) = \frac {1}{1+x^2}.$

To start, note that the Inverse Function Theorem assures us that this derivative actually exists. Recall $\arctan (x) = \theta$ means that $\tan (\theta ) = x$ and $\answer [given]{\frac {-\pi }{2}}< \theta < \answer [given]{\frac {\pi }{2}}$ . Implicitly differentiating with respect $x$ we see $\begin{align*} \tan(\theta) &= x\\ \ddx \tan(\theta) &= \ddx x &\text{Differentiate both sides.}\\ \sec^2(\theta) \cdot \theta' &= 1 &\text{Implicit differentiation.}\\ \theta' &= \frac{1}{\sec^2(\theta)} \end{align*}$

Recall the trig identity: $\sec ^2(\theta )=1+\tan ^2(\theta )$ .

$\begin{align*} \theta' &= \frac{1}{1+\tan^2(\theta)} \end{align*}$

Recall from above: $\tan (\theta ) = x$ .

$\begin{align*} \theta' &= \frac{1}{1+x^2} \end{align*}$

To be completely explicit, $\ddx \theta = \ddx \arctan (x) = \answer [given]{\frac {1}{1+x^2}}.$

Compute: $\ddx \tan ^{-1}(\sqrt {x}) \begin {prompt} = \answer [given]{1/(2\sqrt {x}(1+x))} \end {prompt}$

Finally, we investigate the derivative of arcsecant.

The derivative of arcsecant $\ddx \arcsec (x) = \frac {1}{|x|\sqrt {x^2-1}}\qquad \text {for $|x|>1$.}$

Recall $\arcsec (x) = \theta$ means that $\sec (\theta ) = x$ and $\answer [given]{0}\le \theta \le \answer [given]{\pi }$ with $\theta \ne \answer [given]{\pi /2}$ . Implicitly differentiating with respect $x$ we see $\begin{align*} \sec(\theta) &= x\\ \ddx \sec(\theta) &= \ddx x &\text{Differentiate both sides.}\\ \sec(\theta)\tan(\theta) \cdot \theta' &= 1 &\text{Implicit differentiation.}\\ \theta' &= \frac{1}{\sec(\theta)\tan(\theta)} &\text{Solve for $\theta'$.}\\ \theta' &= \frac{\cos^2(\theta)}{\sin(\theta)}. \end{align*}$

While $\theta ' = \frac {\cos ^2(\theta )}{\sin (\theta )}$ , we need our answer written in terms of $x$ . Since we are assuming that $\sec (\theta ) = \frac {1}{\cos (\theta )}= x,$ we may consider the following triangle:

We may now scale this triangle by a factor of $\frac {1}{x}$ to place it on the unit circle:

From the unit circle above, we see that $\begin{align*} \theta' &= \frac{\cos^2(\theta)}{\sin(\theta)}\\ &= \frac{ \left(\frac{\mathrm{adj}}{\mathrm{hyp}}\right)^2}{\frac{\mathrm{opp}}{\mathrm{hyp}}}\\ &= \frac{\left(\mathrm{adj}\right)^2}{\mathrm{opp}} &\text{Note, $\mathrm{hyp}=1$.}\\ &= \frac{\answer[given]{1/x^2}}{\sqrt{1-1/x^2}}\\ &= \frac{1}{|x|\sqrt{x^2-1}}, \end{align*}$

To be completely explicit, $\ddx \theta = \ddx \arcsec (x) = \frac {1}{|x|\sqrt {x^2-1}}\qquad \text {for $|x|>1$}.$

Compute: $\ddx \sec ^{-1}(3x) \begin {prompt} = \answer [given]{\frac {1}{|x|\sqrt {(3x)^2-1}}} \end {prompt}$

We leave it to you, the reader, to investigate the derivatives of cosine, arccosecant, and arccotangent. However, as a gesture of friendship, we now present you with a list of derivative formulas for inverse trigonometric functions.

The Derivatives of Inverse Trigonometric Functions

$\dd {x} \arcsin (x) = \frac {1}{\sqrt {1-x^2}}$
$\dd {x} \arccos (x) = \frac {-1}{\sqrt {1-x^2}}$
$\dd {x} \arctan (x) = \frac {1}{1+x^2}$
$\dd {x} \arcsec (x) = \frac {1}{|x|\sqrt {x^2-1}}$ for $|x|>1$
$\dd {x} \arccsc (x) = \frac {-1}{|x|\sqrt {x^2-1}}$ for $|x|>1$
$\dd {x} \arccot (x) = \frac {-1}{1+x^2}$