Chapter 13.3, Problem 23E

To determine

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

Answer to Problem 23E

  $f\text{'}\left(a\right)=\frac{1}{{\left(1+a\right)}^2}$

Explanation of Solution

Given information:

The given function and the point is $f\left(x\right)=\frac{x}{1+x}$.

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

  $\begin{array}{l}f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\frac{\left(a+h\right)}{1+\left(a+h\right)}-\frac{a}{\left(1+a\right)}}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\frac{\left(a+h\right)}{1+\left(a+h\right)}-\frac{a}{\left(1+a\right)}}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\frac{\left(a+h\right)\left(1+a\right)-a\left\{1+\left(a+h\right)\right\}}{\left\{1+\left(a+h\right)\right\}\left(1+a\right)}}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{a+a^2+h+ah-a-a^2-ah}{h\left\{1+\left(a+h\right)\right\}\left(1+a\right)}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\cancel{h}}{\cancel{h}\left\{1+\left(a+h\right)\right\}\left(1+a\right)}\\ f\text{'}\left(a\right)=\frac{1}{\left\{1+\left(a+0\right)\right\}\left(1+a\right)}\\ f\text{'}\left(a\right)=\frac{1}{{\left(1+a\right)}^2}\end{array}$

Hence,

  $f\text{'}\left(a\right)=\frac{1}{{\left(1+a\right)}^2}$