Chapter 3.1, Problem 7E

To determine

To calculate: The derivative of the given function at the given value of a using the definition $f^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Answer to Problem 7E

The required value is $\frac{1}{4}$ .

Explanation of Solution

Given information:

The function: $f\left(x\right)=\sqrt{x+1}$ , $a=3$

Calculation:

The given function is $f\left(x\right)=\sqrt{x+1}$ .

Write the given definition using the above function as:

  $f^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{\left(\sqrt{x+1}\right)-\left(\sqrt{a+1}\right)}{x-a}$

Substitute 3 for a in the above limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(3\right)=\underset{x\to 3}{\lim }\frac{\left(\sqrt{x+1}\right)-\left(\sqrt{3+1}\right)}{x-3}\\ =\underset{x\to 3}{\lim }\left[\frac{\sqrt{x+1}-2}{x-3}\right]\\ =\underset{x\to 3}{\lim }\left[\frac{\left(\sqrt{x+1}-2\right)}{\left(x-3\right)}\cdot \frac{\left(\sqrt{x+1}+2\right)}{\left(\sqrt{x+1}+2\right)}\right]\\ =\underset{x\to 3}{\lim }\left[\frac{x+1-4}{\left(x-3\right)\left(\sqrt{x+1}+2\right)}\right]\end{array}$

Simplify the limit further.

  $\begin{array}{c}{f}^{\prime }\left(3\right)=\underset{x\to 3}{\lim }\left[\frac{x+1-4}{\left(x-3\right)\left(\sqrt{x+1}+2\right)}\right]\\ =\underset{x\to 3}{\lim }\left[\frac{\cancel{\left(x-3\right)}}{\cancel{\left(x-3\right)}\left(\sqrt{x+1}+2\right)}\right]\\ =\underset{x\to 3}{\lim }\left[\frac{1}{\sqrt{x+1}+2}\right]\end{array}$

Substitute the 3 for x in the limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(3\right)=\frac{1}{\sqrt{3+1}+2}\\ =\frac{1}{\sqrt{4}+2}\\ =\frac{1}{2+2}\\ =\frac{1}{4}\end{array}$

Hence, the required value is $\frac{1}{4}$ .