Chapter 11, Problem 9CT

To determine

To calculate:

The derivative of the following function:

  $f\left(x\right)=\frac{1}{x+1}$

Answer to Problem 9CT

The derivative of the following function of $f\left(x\right)=\frac{1}{x+1}$ is $f\left(x\right)=-\frac{1}{{\left(x+1\right)}^2}$ .

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{1}{x+1}$

Calculation:

Consider the function:

  $f\left(x\right)=\frac{1}{x+1}$

Derivative with respect to $\frac{d}{dx}$ :

  $\begin{array}{l}\frac{d}{dx}\left[\frac{1}{x+1}\right]=-\frac{\frac{d}{dx}\left[x+1\right]}{{\left(x+1\right)}^2}\\ \frac{d}{dx}\left[\frac{1}{x+1}\right]=-\frac{\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[1\right]}{{\left(x+1\right)}^2}\\ \frac{d}{dx}\left[\frac{1}{x+1}\right]=-\frac{1+0}{{\left(x+1\right)}^2}\\ \frac{d}{dx}\left[\frac{1}{x+1}\right]=-\frac{1}{{\left(x+1\right)}^2}\end{array}$