Question

Find the following for the given functionf:

1. $f\text{'}\left(a\right),$where ais in the domain of f, and
2. $f\text{'}\left(3\right)$and$f\text{'}\left(4\right)$.

.$f\left(x\right)=\frac{x}{x+1}$

Short answer

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

(b) The value of $f\text{'}\left(3\right)=\frac{1}{16}andf\text{'}\left(4\right)=\frac{1}{25}$.

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=\frac{x}{x+1}$

Step 2. Formula used.

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

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

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative at, a we have

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

Therefore, the derivative of the function at a is$f\text{'}\left(a\right)=\frac{1}{{(a+1)}^2}$ .

Step 4. Finding the value of derivative at given point.

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

Let’s find the value of derivative at$a=3anda=4$.

Substitute the value of $a=3anda=4$into the$f\text{'}\left(a\right)=\frac{1}{{(a+1)}^2}$

$\begin{array}{c}f\text{'}\left(3\right)=\frac{1}{{(3+1)}^2}\\ =\frac{1}{16}\\ f\text{'}\left(4\right)=\frac{1}{{(4+1)}^2}\\ =\frac{1}{25}\end{array}$

Therefore, the value of$f\text{'}\left(3\right)=\frac{1}{16}andf\text{'}\left(4\right)=\frac{1}{25}$ .