Chapter 3.6, Problem 44E

  $\left(a\right)$

To determine

To graph: The function $f\left(x\right)=\sin \left({\sin }^{-1}x\right)$ .

Explanation of Solution

Given information:

The equation $f\left(x\right)=\sin \left({\sin }^{-1}x\right)$ .

Graph:

On the coordinate axes marks the points on both the axes in sketch the graph of the function $f\left(x\right)=\sin \left({\sin }^{-1}x\right)$ . The graph obtained is provided below,

Interpretation:

The provided function is composed of inverse trigonometric and trigonometric function. The domain of inverse sine trigonometric function is $\left[-1,1\right]$ . Also the same trigonometric function and its inverse cancel out each other so,

  $\begin{array}{c}f\left(x\right)=\sin \left({\sin }^{-1}x\right)\\ =x\end{array}$

The equation represents a straight line $y=x$ that passes through the origin, but $x\in \left[-1,1\right]$ because of restricted domain of ${\sin }^{-1}\left(x\right)$ .

  $\left(b\right)$

To determine

To graph: The function $g\left(x\right)={\sin }^{-1}\left(\sin x\right)$ .

Explanation of Solution

Given information:

The equation $g\left(x\right)={\sin }^{-1}\left(\sin x\right)$ .

Graph:

On the coordinate axes marks the points on both the axes in sketch the graph of the function $g\left(x\right)={\sin }^{-1}\left(\sin x\right)$ . The graph obtained is provided below,

Interpretation:

The provided function is composed of inverse trigonometric and trigonometric function. The domain of sine trigonometric function is $ℝ$ . Therefore, the function $g\left(x\right)={\sin }^{-1}\left(\sin x\right)$ is defined for all real values of x . Also the range of sine trigonometric function is $\left[-1,1\right]$ so the $\arcsin$ of it is possible since domain of inverse sine trigonometric function is $\left[-1,1\right]$ . The value obtained in graph oscillates between $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ which is the range of inverse sine trigonometric function.

  $\left(c\right)$

To determine

To show: The value of the expression $g\text{'}\left(x\right)=\frac{\cos x}{\left|\cos x\right|}$ .

Explanation of Solution

Given information:

The function $g\left(x\right)={\sin }^{-1}\left(\sin x\right)$ .

Formula used:

Derivation of ${\sin }^{-1}\left(x\right)$ is $\frac{d}{dx}{\sin }^{-1}\left(x\right)=\frac{1}{\sqrt{1-x^2}}$ .

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Calculation:

Consider the provided function $g\left(x\right)={\sin }^{-1}\left(\sin x\right)$ .

Recall that thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ . Also derivation of ${\sin }^{-1}\left(x\right)$ is $\frac{d}{dx}{\sin }^{-1}\left(x\right)=\frac{1}{\sqrt{1-x^2}}$ .

Apply it,

  $g\text{'}\left(x\right)=\frac{1}{\sqrt{1-{\sin }^{2}x}}\frac{d}{dx}\left(\sin x\right)$

Apply the Pythagorean identity, ${\sin }^{2}x+{\cos }^{2}x=1$ ,

  $\begin{array}{c}g\text{'}\left(x\right)=\frac{1}{\sqrt{1-{\sin }^{2}x}}\frac{d}{dx}\left(\sin x\right)\\ =\frac{1}{\sqrt{1-{\sin }^{2}x}}\left(\cos x\right)\\ =\frac{\cos x}{\left|\cos x\right|}\end{array}$

Thus, it is shown that the value of the expression $g\text{'}\left(x\right)=\frac{\cos x}{\left|\cos x\right|}$ .

  $\left(d\right)$

To determine

To calculate: The derivative of the function $h\left(x\right)={\cos }^{-1}\left(\sin x\right)$ . Also sketch the graph of the function.

Answer to Problem 44E

The derivative of the function is $h\text{'}\left(x\right)=\frac{-\cos x}{\left|\cos x\right|}$ . The graph of the function is provided below,

Explanation of Solution

Given information:

The function $h\left(x\right)={\cos }^{-1}\left(\sin x\right)$ .

Formula used:

Derivation of ${\cos }^{-1}\left(x\right)$ is $\frac{d}{dx}{\cos }^{-1}\left(x\right)=-\frac{1}{\sqrt{1-x^2}}$ .

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Calculation:

Consider the provided function $h\left(x\right)={\cos }^{-1}\left(\sin x\right)$ .

Recall that thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ . Also derivation of ${\cos }^{-1}\left(x\right)$ is $\frac{d}{dx}{\cos }^{-1}\left(x\right)=-\frac{1}{\sqrt{1-x^2}}$ .

Apply it,

  $h\text{'}\left(x\right)=-\frac{1}{\sqrt{1-{\sin }^{2}x}}\frac{d}{dx}\left(\sin x\right)$

Apply the Pythagorean identity, ${\sin }^{2}x+{\cos }^{2}x=1$ ,

  $\begin{array}{c}h\text{'}\left(x\right)=-\frac{1}{\sqrt{1-{\sin }^{2}x}}\frac{d}{dx}\left(\sin x\right)\\ =-\frac{1}{\sqrt{1-{\sin }^{2}x}}\left(\cos x\right)\\ =\frac{-\cos x}{\left|\cos x\right|}\end{array}$

Thus, the derivative of the function is $h\text{'}\left(x\right)=\frac{-\cos x}{\left|\cos x\right|}$ .

On the coordinate axes marks the points on both the axes in sketch the graph of the function $h\left(x\right)={\cos }^{-1}\left(\sin x\right)$ . The graph obtained is provided below,

The range of ${\cos }^{-1}\left(x\right)$ is $\left[0,\pi \right]$ , so for all real values of x the output falls between the range.