Chapter 4.4, Problem 40E

To determine

To find: the first derivative of the given function.

Answer to Problem 40E

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

Explanation of Solution

Given information:

  $f\left(x\right)=\ln \left(x^2+1\right)$

Calculation:

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

The domain would be stated as for all real numbers.

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