Chapter 4.4, Problem 41E

To determine

To find the first derivative of the given function.

Answer to Problem 41E

The first derivative of the function $f\left(x\right)={\log }_2\left(3x+1\right)$ is $f^{\prime }\left(x\right)=\frac{3}{\ln \left(2\right)\left(3x+1\right)}$ and the state of domain is $x>-\frac{1}{3}$ .

Explanation of Solution

Given information:

  $f\left(x\right)={\log }_2\left(3x+1\right)$

Calculation:

  $\begin{array}{l}{f}^{\prime }\left(x\right)={\left({\log }_2\left(3x+1\right)\right)}^{\prime }\\ f^{\prime }\left(x\right)=\frac{1}{\ln \left(2\right)}{\left(\ln \left(u\right)\right)}^{\prime }{\left(3x+1\right)}^{\prime }\to \text{by using the chain rule}\\ f^{\prime }\left(x\right)=\frac{1}{\ln \left(2\right)}\times \frac{1}{u}\times 3\\ f^{\prime }\left(x\right)=\frac{1}{\ln \left(2\right)}\times \frac{1}{3x+1}\times 3\\ f^{\prime }\left(x\right)=\frac{3}{\ln \left(2\right)\left(3x+1\right)}\end{array}$

The domain would be stated as $x>-\frac{1}{3}$ .

Thus we can find the first derivative of the given function and state the domain.