\[ \ddx \sin ^{-1}(x) \begin{prompt} = \answer [given]{1/\sqrt {1-x^2}} \end{prompt} \]
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
![[Picture]](digInDerivativesOfInverseTrigonometricFunctions.online-9d9b6487498fb641b2a4dde40eeec041.svg)
From the unit
circle above, we see that
\[ \arcsin (x) = \theta \]
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: \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:
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 to \(x\) we obtain that
\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 ![[Picture]](digInDerivativesOfInverseTrigonometricFunctions.online-7135656ab6de514694a8afac77386895.svg)
We
may now scale this triangle by a factor of \(\frac {1}{x}\) to place it on the unit circle:
![[Picture]](digInDerivativesOfInverseTrigonometricFunctions.online-86b150d55c74e9d9346c0e29d3b2c084.svg)
From the
unit circle above, we see that
\[ \arcsec (x) = \theta \]
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: \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 \(|x|>1\)
- \(\dd {x} \arccot (x) = \frac {-1}{1+x^2}\)
Let’s get some practice using these new shortcut-formulas.