Chapter 3.6, Problem 31E

To determine

To calculate: The derivative of the function $g\left(x\right)={\cos }^{-1}\left(3-2x\right)$ . Also evaluate the domain of the function and its derivative.

Answer to Problem 31E

The derivative of the function is $g\text{'}\left(x\right)=\frac{2}{\sqrt{1-{\left(3-2x\right)}^2}}$ . The domain of the function is $x\in \left[1,2\right]$ and domain of derivative of the function is $x\in \left(1,2\right)$ .

Explanation of Solution

Given information:

The function $g\left(x\right)={\cos }^{-1}\left(3-2x\right)$ .

Formula used:

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

Derivative of inverse trigonometric function $\frac{d}{dx}\left(\arccos \left(x\right)\right)=\frac{-1}{\sqrt{1-x^2}}$ .

Calculation:

Consider the function $g\left(x\right)={\cos }^{-1}\left(3-2x\right)$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(g\left(x\right)\right)=\frac{d}{dx}\left({\cos }^{-1}\left(3-2x\right)\right)\\ g\text{'}\left(x\right)=\frac{d}{dx}\left({\cos }^{-1}\left(3-2x\right)\right)\end{array}$

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

Derivative of inverse trigonometric function $\frac{d}{dx}\left(\arccos \left(x\right)\right)=\frac{-1}{\sqrt{1-x^2}}$ .

Apply it,

  $\begin{array}{c}g\text{'}\left(x\right)=\frac{-1}{\sqrt{1-{\left(3-2x\right)}^2}}\frac{d}{dx}\left(3-2x\right)\\ g\text{'}\left(x\right)=\frac{-1\left(-2\right)}{\sqrt{1-{\left(3-2x\right)}^2}}\\ g\text{'}\left(x\right)=\frac{2}{\sqrt{1-{\left(3-2x\right)}^2}}\end{array}$

Thus, the derivative of the function is $g\text{'}\left(x\right)=\frac{2}{\sqrt{1-{\left(3-2x\right)}^2}}$ .

Consider the function, $g\text{'}\left(x\right)=\frac{2}{\sqrt{1-{\left(3-2x\right)}^2}}$ .

Recall that domain of the function is the set of values of independent variable for which function is defined.

The function of the form $\frac{p}{q}$ fraction are defined when $q\ne 0$ that is denominator should not be equal to 0.Also, the square root sign is defined for positive values.

The provided expression $g\text{'}\left(x\right)=\frac{2}{\sqrt{1-{\left(3-2x\right)}^2}}$ is defined for real values that is, $1-{\left(3-2x\right)}^2>0$ , therefore,

  $\begin{array}{c}1>{\left(3-2x\right)}^2\\ -1<3-2x<1\\ -4<-2x<-2\\ 1<x<2\end{array}$

Therefore, domain is $x\in \left(1,2\right)$ for function to be defined.

Consider the function, $g\left(x\right)={\cos }^{-1}\left(3-2x\right)$ .

Recall that domain of the function is the set of values of independent variable for which function is defined.

The domain of inverse cosine trigonometric function is $\left[-1,1\right]$ . Therefore,

  $\begin{array}{c}-1\le 3-2x\le 1\\ -4\le -2x\le -2\\ -2\le x\le -1\\ 1\le x\le 2\end{array}$

Therefore, domain is $x\in \left[1,2\right]$ for function to be defined.

Hence, the derivative of the function is $g\text{'}\left(x\right)=\frac{2}{\sqrt{1-{\left(3-2x\right)}^2}}$ . The domain of the function is $x\in \left[1,2\right]$ and domain of derivative of the function is $x\in \left(1,2\right)$ .