Q. 51 - Calculus

Question

Use the definition of the derivative to find $f\text{'}$ for each function $f$ in Exercises 39-54

$f\left(x\right)=\frac{x-1}{x+3}$

Step-by-Step Solution

Answer

The value of $f\text{'}\left(x\right)=\frac{4}{{(x+3)}^2}$

Step 1. Given information

The given function$f\left(x\right)=\frac{x-1}{x+3}$

Step 2. Finding the value of f '

We know that $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}..............\left(1\right)$

Given $f\left(x\right)=\frac{x-1}{x+3}$ then

$f(x+h)=\frac{x+h-1}{x+h+3}$

Putting these values in (1)

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x+h-1}{x+h+3}}-{\displaystyle \frac{x-1}{x+3}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{(x+3)(x+h-1)-(x-1)(x+h+3)}{(x+h+3)(x+3)}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{(x^2+xh-x+3x+3h-3)-(x^2+xh+3x-x-h-3)}{(x+h+3)(x+3)}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x^2+xh-x+3x+3h-3-x^2-xh-3x+x+h+3}{(x+h+3)(x+3)}}}{h} \\ =\underset{h\to 0}{\lim }\frac{\frac{3h+h}{(x+h+3)(x+3)}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{h(3+1)}{(x+h+3)(x+3)}}}{h} \\ =\underset{h\to 0}{\lim }\frac{4h}{(x+h+3)(x+3)}\times \frac{1}{h} \\ =\underset{h\to 0}{\lim }\frac{4}{(x+h+3)(x+3)} \end{gathered}$$

Putting $h=0$

        $$\begin{gathered} =\frac{4}{(x+0+3)(x+3)} \\ =\frac{4}{(x+3)(x+3)} \\ =\frac{4}{{(x+3)}^2} \end{gathered}$$

Hence,the value of $f\text{'}\left(x\right)=\frac{4}{{(x+3)}^2}$