Chapter 4.4, Problem 38E

To determine

To find the first derivative of the given function.

Answer to Problem 38E

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

Explanation of Solution

Given information:

  $f\left(x\right)=\ln \left(2x+2\right)$

Calculation:

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

The domain would be stated as $x>-1$ .

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