Derivatives of inverse trigonometric functions

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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 $|x|>1$
$\dd {x} \arccot (x) = \frac {-1}{1+x^2}$