Question

Use (a) the h→0 definition of the derivative and then (b) the z→c definition of the derivative to find f'(c) for each function f and value x = c.

$f\left(x\right)=\frac{x-1}{x+3},x=2$

Short answer

(a) $f^{\text{'}}\left(c\right)=\frac{4}{25}$

(b)$f^{\text{'}}\left(c\right)=\frac{4}{25}$

Step-by-step solution

Part (a) Step 1. Given information.

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

We have to find$f^{\text{'}}\left(c\right)$at x=2

Part (a) Step 2. Find the f'(c)

We have to find the derivative of the function using h→0 definition,

Therefore,

$\begin{array}{r}\underset{h\to 0}{lim} \frac{f(2+h)-f\left(2\right)}{h}=\underset{h\to 0}{lim} \frac{\frac{2+h-1}{2+h+3}-\frac{2-1}{2+3}}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{1+h}{5+h}-\frac{1}{5}}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{5(1+h)-1(5+h)}{5(5+h)}}{h}\\ =\underset{h\to 0}{lim} \frac{4h}{5h(5+h)}\\ =\underset{h\to 0}{lim} \frac{4}{5(5+h)}\\ =\frac{4}{25}\end{array}$

Part (b) Step 1. Find f'(c)

Find the derivate of the function using $x\to 2$definition,

$\begin{array}{r}\underset{x\to 2}{lim} \frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2}{lim} \frac{\frac{x-1}{x+3}-\frac{2-1}{2+3}}{x-2}\\ =\underset{x\to 2}{lim} \frac{\frac{x-1}{x+3}-\frac{1}{5}}{x-2}\\ =\underset{x\to 2}{lim} \frac{\frac{5(x-1)-(x+3)}{5(x+3)}}{x-2}\\ =\underset{x\to 2}{lim} \frac{4x-8}{5(x-2)(x+3)}\\ =\underset{x\to 2}{lim} \frac{4(x-2)}{5(x-2)(x+3)}\\ =\underset{x\to 2}{lim} \frac{4}{5(x+3)}\\ =\frac{4}{25}\end{array}$