Chapter 13.3, Problem 21E

To determine

To find : $f\text{'}\left(a\right)$where a is in the domain of f.

Answer to Problem 21E

  $f\text{'}\left(a\right)=2a+2$

Explanation of Solution

Given information:

The given function and the point is $f\left(x\right)=x^2+2x$.

Calculation :

The derivative of a function $f$ at a number $a$ , denoted by $f\text{'}\left(a\right)$ ,is

  $f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$

provided this limit exist.

According to definition, $f\text{'}\left(a\right)$ is

  $\begin{array}{l}f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{{\left(a+h\right)}^2+2\left(a+h\right)-\left(a^2+2a\right)}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{a^2+2ah+h^2+2a+2h-a^2-2a}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{2ah+h^2+2h}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\cancel{h}\left(2a+h+2\right)}{\cancel{h}}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\left(2a+h+2\right)\\ f\text{'}\left(a\right)=2a+2\end{array}$

Hence,

  $f\text{'}\left(a\right)=2a+2$