Chapter 4, Section 7

4.7 Derivatives of Inverse Trigonometric Functions

Goals

The goal of this section is twofold: on one hand, to quickly review the definitions of the inverse trigonometric functions, and, on the other hand, to apply the Chain Rule to compute their derivatives. In fact, the process to compute the derivatives is similar for all of them, and consists of four basic steps.

4.7.1 Derivative of arcsine

Basic Properties of arcsinine

Recall that in order to define \(y =\arcsin x\), it is necessary to restrict the domain of \(y = \sin x\) to an interval where it is one-to-one. This interval is usually chosen to be \(\left [-\frac{\pi}{2}, \frac{\pi}{2}\right ]\) (see Figure 4.10(a)). Thus, the domain and range of \(y=\arcsin x\) are, respectively,
\[D_{\arcsin}=[-1,1] \;\;\; R_{\arcsin}=\left [-\frac{\pi}{2}, \frac{\pi}{2}\right ].\]
Furthermore,
\[\sin(\arcsin x)=x \;\; \text{for all} \; x\in D_{\arcsin},\;\;\; \arcsin(\sin x)=x \;\; \text{for all} \; x \in\left [-\frac{\pi}{2}, \frac{\pi}{2}\right ].\]

Figure 4.10 (a)

Figure 4.10 (b)

Figure 4.10 (c)

Derivative of arcsine

Proposition 1. Let \(y=\arcsin x\), then
\begin{equation}\label{ch04-07-eq01}
\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, \;\;\; \forall x\in (-1, 1).\tag{4.5}
\end{equation}

Proof.

Step 1. Let \(x\in (-1,1)\). Then
\[y = \arcsin x \Longleftrightarrow \sin y = x \text{ and } y\in\left (-\frac{\pi}{2}, \frac{\pi}{2}\right ).\]

Step 2. Applying the Chain Rule
\[\frac{d}{dx}\sin y = \frac{dx }{dx}\;\;\Longrightarrow \;\; \cos y\, y’=1 \;\;\Longleftrightarrow\;\; y’=\frac{1}{\sin y}.\]
Step 3. Applying the Pythagorean identity, and using the fact that \(\cos y > 0\) for all \(y \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
\[\sin y = x \Longrightarrow \cos^2 y = 1-\sin^2 y = 1-x^2\;\; \Longrightarrow \;\; \cos y = \sqrt{1-x^2}.\]
Step 4. Conclusion
\[y’=\frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}.\]

Example 1.

Find the domain and the derivative of the function (a) \(y = \arcsin ax\), \(a \in \mathbb{R}\) nonzero constant;
(b) \(y = \arcsin^3\sqrt{x}\); (c) \(y = \cot (\arcsin \sqrt{x-1})\).

Solution. (a) \(D=\left [-\frac{1}{a},\frac{1}{a}\right ] \). Letting \(f(x)= \arcsin x\) and \(g(x) = ax\), we have
\[y = (f\circ g)(x) \;\; \Longrightarrow \;\; y’ = f'(g(x))g'(x) = \frac{1}{\sqrt{1-(g(x))^2}}\cdot a= \frac{a}{\sqrt{1-a^2x^2}}.\]

(b) \(x\in D \Longleftrightarrow \sqrt{x} \leq 1 \Longleftrightarrow x\in [0,1]\). To compute \(y’\) apply the Chain Rule twice, first with \(f(x)= x^3\) and \(g(x) = \arcsin \sqrt x\), and then again to \(g(x) = \arcsin \sqrt x\)
\[y’ = 3\left (\arcsin \sqrt{x}\right )^2g'(x) = 3 \arcsin^2 \sqrt{x} \cdot \left (\frac{1}{\sqrt{1-\left (\sqrt{x}\,\right )^2}}\right )\cdot\frac{1}{2\sqrt{x}} =
\frac{3 \arcsin^2 \sqrt{x}}{2 \sqrt{x-x^2}}.\]

(c) Verify that \(D= (1,2)\). Applying the Chain Rule twice
\[
y’= -\csc^2\arcsin\sqrt{x-1}\cdot \left (\frac{1}{\sqrt{1-\left (\sqrt{x-1}\,\right)^2}}\right )\cdot\frac{1}{2\sqrt{x-1}}
= -\frac{\csc^2\arcsin\sqrt{x-1}}{2\sqrt{(2-x)(x-1)}}.\]
The expression \(\csc^2\arcsin\sqrt{x-1}\) can be simplified as follows:
\[\csc^2\arcsin\sqrt{x-1} = \frac{1}{\left (\sin(\arcsin \sqrt{x-1}\,)\right )^2} = \frac{1}{x-1}.\]
Thus
\[y’ = -\frac{1}{2(x-1)\sqrt{(2-x)(x-1)}}.\]

Example 2.
Given \(\displaystyle h(x) =\arcsin \frac{x+1}{x-1}\), (a) find \(D_h\); (b) find and simplify \(h’\); (c) show that if the \(x\)-coordinate of the point on the graph of \(h\) is \(x=-4\), then the its tangent line is parallel to the line \(10x+y = 7\).

Solution. (a) Let \(f(x) = \arcsin x\) and \(\displaystyle g(x) = \frac{x+1}{x-1}\).
\[x\in D_h \Longleftrightarrow x\in D_g \;\; \text{and} \;\; g(x)\in D_f \Longleftrightarrow -1\leq \frac{x+1}{x-1} \leq 1.\]
Thus, the following two inequalities must be satisfied simultaneously.
\[-1\leq \frac{x+1}{x-1} \Longleftrightarrow 0 \leq 1 + \frac{x+1}{x-1} \Longleftrightarrow 0 \leq \frac{2x}{x-1} \underset{?}{\Longleftrightarrow} x\in (-\infty, 0]\cup (1, \infty),\]
and
\[\frac{x+1}{x-1} \leq 1 \Longleftrightarrow \frac{x+1}{x-1}-1 \leq 0 \Longleftrightarrow \frac{2}{x-1} \leq 0 \Longleftrightarrow x\in (-\infty, 1).\]
Hence
\[ D_h = \left ( (-\infty, 0]\cup (1, \infty)\right )\cap (-\infty, 1) \Longleftrightarrow D_h= (-\infty, 0].\]
(b) Let, \(x\in D_h\), then, by the Chain Rule

\begin{align*}
h'(x) &= f'(g(x))g'(x) \underset{?}{=} \frac{1}{\sqrt{1-\left (\frac{x+1}{x-1}\right )^2}}\cdot \frac{-2}{( x-1)^2}
= -\frac{1}{\sqrt{ \frac{(x-1)^2-(x+1)^2}{(x-1)^2}}}\cdot \frac{2}{( x-1)^2} =\\
&= -\frac{\sqrt{(x-1)^2}}{\sqrt{ -4x}}\cdot \frac{2}{( x-1)^2}
\underset{?}{=} \frac{1}{\sqrt{-x}(x-1)}, \;\; x\in D_h, \;\; x\neq 0.
\end{align*}

(c) Since the line \(10x+y = 7\) has slope \(\displaystyle m =- \frac{1}{10}\), for a parallel tangent line
\[h'(x) = -\frac{1}{10}\;\; \Longleftrightarrow \;\; \frac{1}{\sqrt{-x}(x-1)}= -\frac{1}{10} \;\; \Longleftrightarrow \;\; \sqrt{-x}(1-x) = 10.\]
Substituting \(x=-4\), \(\sqrt{-(-4)}(1-(-4)) = 2\cdot 5 = 10\).

4.7.2 Derivative of Arccosine

Basic Properties of Arccosine

Recall that in order to define \(y =\arccos x\), it is necessary to restrict the domain of \(y = \cos x\) to an interval where it is one-to-one. This interval is usually chosen to be \([0, \pi]\) (see Figure 4.11). Thus, the domain and range of \(y=\arccos x\) are, respectively,
\[D_{\arccos}=[-1,1] \;\;\; R_{\arccos}=[0,\pi].\]
Furthermore,
\[\cos(\arccos x)=x \;\; \text{for all} \; x\in D_{\arccos},\;\;\; \arccos(\cos x)=x \;\; \text{for all} \; x \in [0,\pi].\]

Figure 4.11

Derivative of Arccosine

Proposition 2. Let \(y=\arccos x\), then

\begin{equation}\label{ch04-07-eq02}
\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}, \;\;\; \forall x\in (-1, 1).\tag{4.6}
\end{equation}

On Your Own 6.
Prove Proposition 2.

4.7.3 Derivative of Arctangent

Basic Properties of Arctangent

To define \(y =\arctan x\), \(y = \tan x\) is usually restricted \( x\in \left (-\frac{\pi}{2}, \frac{\pi}{2}\right )\) (see Figure 4.12(a)). Thus, the domain and range of \(y=\arctan x\) are, respectively,
\[D_{\arctan}=(-\infty, \infty) \;\;\; R_{\arctan}=\left (-\frac{\pi}{2}, \frac{\pi}{2}\right ).\]
Furthermore,
\[\tan(\arctan x)=x \;\; \text{for all} \; x\in D_{\arctan},\;\;\; \arctan(\tan x)=x \;\; \text{for all} \; x \in\left (-\frac{\pi}{2}, \frac{\pi}{2}\right ).\]

Figure 4.12 (a)

Figure 4.12 (b)

Figure 4.12 (c)

Derivative of arctangent

Proposition 3.
Let \(y=\arctan x\), then
\begin{equation}\label{ch04-07-eq03}
\frac{dy}{dx} = \frac{1}{ 1+x^2}, \;\;\; \forall x\in (-\infty, \infty).\tag{4.7}
\end{equation}

Proof.

Step 1. Let \(x\in (-1,1)\). Then
\[y = \arctan x \Longleftrightarrow \tan y = x \text{ and } y\in\left (-\frac{\pi}{2}, \frac{\pi}{2}\right ).\]

Step 2. Applying the Chain Rule
\[\frac{d}{dx}\tan y = \frac{dx }{dx}\;\;\Longrightarrow \;\; \sec^2 y\, y’=1 \;\;\Longleftrightarrow\;\; y’=\frac{1}{\tan y}.\]

Step 3. Since \(\sec^2x= 1+\tan^2x\), we have
\[\tan y = x \Longrightarrow \sec^2 y = 1+\tan^2 y = 1+x^2.\]

Step 4. Conclusion
\[y’=\frac{1}{\sec^2 y} = \frac{1}{1+x^2}.\]

Example 1.
Find the derivative of the function \(F(x)=\arctan^3 \left(\cot 2x\right)\).}

Solution. Here \(u=g(x)=\arctan \left(\cot 2x\right)\) and \(y=f(u)=u^3\) \(\Longrightarrow \)

\[\begin{array}{ccccc}
x & \longmapsto & u=\arctan \left(\cot 2x\right) & \longmapsto & y=u^3 \\
& g'(x)=(*) & & f'(u)=3u^2 & \\
\end{array}, \]

to compute \((*)\) apply the chain rule with \(u = A(x)= \cot 2x \), and \(y = B(u) = \arctan u\Longrightarrow \)

\[\begin{array}{ccccc}
x & \longmapsto & u= \cot 2x & \longmapsto & y=\arctan u \\
& A(x)=-2\csc^22x & & B'(u)=\frac{1}{1+u^2} & \\
\end{array}, \]
Hence
\[ (B\circ A)'(x) = A'(x)B'(u) = -2\csc^22x \frac{1}{1+u^2} \Longrightarrow \]
\[\;\;\;\;\Longrightarrow (B\circ A)'(x)= B'(A(x)) A'(x)=-2\csc^22x \frac{1}{1+\cot^22x}.\;\;\;\; (*)\]
It follows that
\[F'(x) = -2\csc^22x \frac{1}{1+\cot^22x}\left (3\arctan^2 \left(\cot 2x\right)\right )= -6 \frac{\csc^22x }{1+\cot^22x} \arctan^2 \left(\cot 2x\right) . \]

4.7.4 Derivative of Arcsecant

Basic Properties of Arcsecant

To define \(y =\text{arcsec} \,x\), a restriction of \(y = \sec x\) is \( x \in \left [0,\frac{\pi}{2}\right )\cup \left (\frac{\pi}{2}, \pi\right ]\) (see Figure 4.13(a)). Thus, the domain and range of \(y=\text{arcsec} \, x\) are, respectively,
\[D_{\text{arcsec}}=(-\infty, -1]\cup [1, \infty) \;\;\; R_{\text{arcsec}}=\left [0,\frac{\pi}{2}\right )\cup \left (\frac{\pi}{2}, \pi\right ].\]
Furthermore,
\[\sec(\text{arcsec}\, x)=x \;\; \text{for all} \; x\in (-\infty, -1]\cup [1, \infty),\;\;\; \text{arcsec}(\sec x)=x \;\; \text{for all} \; x \in\left [0,\frac{\pi}{2}\right )\cup \left (\frac{\pi}{2}, \pi\right ].\]

Figure 4.13 (a)

Figure 4.13 (b)

Figure 4.13 (c)

Derivative of arctangent

Proposition 4. Let \(y={\rm{arcsec}}\, x\), then

Proof.
Step 1. Let \(x\in D_{\text{arcsec}}\). Then
\[y = \text{arcsec}\, x \Longleftrightarrow \sec y = x \text{ and } y\in\left [0,\frac{\pi}{2}\right )\cup \left (\frac{\pi}{2}, \pi\right ].\]
Step 2. Applying the Chain Rule
\[\frac{d}{dx}\sec y = \frac{dx }{dx}\;\;\Longrightarrow \;\; \sec y \tan y \, y’=1 \;\;\Longleftrightarrow\;\; y’=\frac{1}{\sec y\tan y }.\]
Step 3.
\[\sec y = x \Longrightarrow \tan^2 y = \sec^2 y -1 = x^2-1 \Longrightarrow \tan y = \pm \sqrt{x^2-1}.\]

Step 4. Thus
\[y’=\frac{1}{\sec y\tan y }= \frac{1}{\pm x \sqrt{x^2-1}}.\]
However, note that if \(x \in (1,\infty)\), then \( y =\sec^{-1} x \in \left [0,\frac{\pi}{2}\right )\), then not only \(\sec y > 0\), but also \(tan y > 0\). Thus, the \((+)\)-sign applies.

If \(x \in (-\infty, -1)\), then \( y =\sec^{-1} x \in \left (\frac{\pi}{2}, \pi \right]\), then \(\sec y < 0\) and \(\tan y < 0\). Thus, the \((-)\)-sign applies. But, since \(x<0\), we can replace \(-x = |x|\), and we obtain \[y’=\frac{1}{\sec y\tan y }= \frac{1}{|x| \sqrt{x^2-1}}.\]

Example 1. Given the function \(F(x)=\sin^3 \left(\sec^{-1}( 2x)\right)\), (a) find its domain; (b) find and simplify its derivative. }
Solution. (a) Since \(D_{\sin}=\mathbb{R}\), \(\displaystyle x \in D_F \underset{?}{\Longleftrightarrow } x \in \left (-\infty, -\frac{1}{2}\right ] \cup \left [\frac{1}{2}, \infty \right ) \).
(b) Let \(G(x)= x^3\) and \(H(x) = \sin \left(\sec^{-1}( 2x)\right)\), then \(F= G\circ H\), and by the Chain Rule \[F'(x)= G'(H(x))H'(x) = 3(H(x))^2H'(x) = 3\sin^2 \left(\sec^{-1}( 2x)\right) H'(x).\] Let now \(g(x) = \sin x\) and \(h(x) = \text{arcsec}\, (2x)\), so that \(H= g\circ h\). Applying the Chain Rule again \[H'(x) = g'(h(x))h'(x) = \cos (h(x)h'(x) = \cos \left (\text{arcsec}\, (2x)\right ) h'(x).\] Finally, applying the Chain Rule to \(h\), we obtain (verify this) \[F'(x) = 3\sin^2 \left(\sec^{-1}( 2x)\right) \cos \left (\text{arcsec}\, (2x)\right ) \frac{1}{|x|\sqrt{4x^2-1}}.\] To simplify this last expression, note that \[ \cos \left (\text{arcsec}\, (2x)\right ) = \frac{1}{ \sec \left (\text{arcsec}\, (2x)\right )} = \frac{1}{2x},\] and \[\sin^2 \left(\sec^{-1}( 2x)\right) = 1- \cos^2 \left(\sec^{-1}( 2x)\right) \underset{?}{=} 1 – \frac{1}{4x^2} =\frac{4x^2 – 1}{4x^2} .\] Hence \[F'(x) = 3\frac{4x^2 – 1}{4x^2} \cdot \frac{1}{2x}\cdot \frac{1}{|x|\sqrt{4x^2-1}} \underset{?}{=} \frac{3\sqrt{4x^2-1}}{8x^3|x|}. \]

4.7.6 Solution to On Your Own 6

Step 1. Let \(x\in (-1,1)\). Then \(y = \arccos x \Longleftrightarrow \cos y = x \text{ and } y\in (0, \pi)\).
Step 2. Applying the Chain Rule \(\displaystyle \frac{d}{dx}\cos y = \frac{dx }{dx}\;\;\Longrightarrow \;\; -\sin y y’=1 \;\;\Longleftrightarrow\;\; y’=-\frac{1}{\sin y}\).
Step 3. Applying the Pythagorean identity, and using the fact that \(\sin y > 0\) for all \(y \in (0,\pi)\):
\(\displaystyle \cos y = x \Longrightarrow \sin^2 y = 1-\cos^2 y = 1-x^2\;\; \Longrightarrow \;\; \sin y = \sqrt{1-x^2}\).

Step 4. Conclusion
\(\displaystyle y’=-\frac{1}{\sin y} = -\frac{1}{\sqrt{1-x^2}}\).