Chapter 2, Problem 76RE

To determine

To find: The relationship between two function $y_1={\cos }^{-1}\left(\cos x\right)\text{ and }{y}_2=\frac{\left|\sin x\right|}{\sin x}$

Answer to Problem 76RE

  $y_2=\frac{\left|\sin x\right|}{\sin x}$ is derivative of the function $y_1={\cos }^{-1}\left(\cos x\right)$ .

Explanation of Solution

Given Information:

  $y_1={\cos }^{-1}\left(\cos x\right)\text{ and }{y}_2=\frac{\left|\sin x\right|}{\sin x}$

Calculation:

Using graphing calculator, graph both functions on same coordinate plane.

  

The period of the both functions is $2\pi$ .

Noticed the graph, If $y_1$ increases the value of $y_2=1$ or positive on interval $\left[0,\pi \right]$ and if $y_1$ decreases the value of $y_2=-1$ or negative on interval $\left[\pi ,2\pi \right]$ .

On interval $\left[0,\pi \right]$:

If $y_2$ is positive then $y_1$ increases.

If $y_2$ is negative then $y_1$ decrease. It means, $y_2$ is derivative of function $y_1$ .

Conclusion:

  $y_2$ is derivative of the function $y_1$ .

To determine

To find: The function which has same relation as part (a) with $y_1={\sin }^{-1}\left(\sin x\right)$ .

Answer to Problem 76RE

  $y_2=\frac{\cos x}{\left|\cos x\right|}$

Explanation of Solution

Given Information:

  $y_1={\sin }^{-1}\left(\sin x\right)$

Calculation:

In part (a), $y_2$ was derivative of function $y_1$ .

To find the derivative of $y_1={\sin }^{-1}\left(\sin x\right)$ ,

  $\begin{array}{l}{y}_2=\frac{\cos x}{\sqrt{1-{\sin }^{2}x}}\left[1-{\sin }^{2}x={\cos }^{2}x\right]\\ y_2=\frac{\cos x}{\sqrt{{\cos }^{2}x}}\left[\sqrt{a^2}=\left|a\right|\right]\\ y_2=\frac{\cos x}{\left|\cos x\right|}\end{array}$

Thus, the function is $y_2=\frac{\cos x}{\left|\cos x\right|}$ .