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

\[ \frac {d}{dx} \arcsin (x) = \frac {1}{\sqrt {1-x^2}}. \]

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\\ \frac {d}{dx} \sin (\theta ) &= \frac {d}{dx} 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,

\[ \frac {d}{dx} \theta = \frac {d}{dx} \arcsin (x) = \answer [given]{\frac {1}{\sqrt {1-x^2}}}. \]

Compute:

\[ \frac {d}{dx} \sin ^{-1}(x) \begin{prompt} = \answer [given]{1/\sqrt {1-x^2}} \end{prompt} \]

We can do something similar with arctangent.

The derivative of arctangent

\[ \frac {d}{dx} \arctan (x) = \frac {1}{1+x^2}. \]

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 to $x$ we obtain that \begin{align*} \tan (\theta ) &= x\\ \frac {d}{dx} \tan (\theta ) &= \frac {d}{dx} 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,

\[ \frac {d}{dx} \theta = \frac {d}{dx} \arctan (x) = \answer [given]{\frac {1}{1+x^2}}. \]

Compute:

\[ \frac {d}{dx} \tan ^{-1}(\sqrt {x}) \begin{prompt} = \answer [given]{1/(2\sqrt {x}(1+x))} \end{prompt} \]

Finally, we investigate the derivative of arcsecant.

The derivative of arcsecant

\[ \frac {d}{dx} \text {arcsec(x)} = \frac {1}{|x|\sqrt {x^2-1}}\qquad \text {for $|x|>1$.} \]

Recall

\[ \text {arcsec(x)} = \theta \]

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\\ \frac {d}{dx} \sec (\theta ) &= \frac {d}{dx} 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,

\[ \frac {d}{dx} \theta = \frac {d}{dx} \text {arcsec(x)} = \frac {1}{|x|\sqrt {x^2-1}}\qquad \text {for $|x|>1$}. \]

Compute:

\[ \frac {d}{dx} \sec ^{-1}(3x) \begin{prompt} = \answer [given]{\frac {1}{|x|\sqrt {(3x)^2-1}}} \end{prompt} \]

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

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

2025-01-06 19:55:39