Question

Use the definition of the derivative to find $f\text{'}$for each function $f$in Exercises39-54

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

Short answer

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

Step-by-step solution

Step 1. Given information

The given function$f\left(x\right)=\frac{2}{1-x}$

Step 2. Finding the value of f'(x)

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 $f\left(x\right)=\frac{2}{1-x}$then

$$\begin{gathered} f(x+h)=\frac{2}{1-(x+h)} \\ =\frac{2}{1-x-h} \end{gathered}$$

Putting these values in(1)

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

Put $h=0$

$$\begin{gathered} =\frac{2}{(1-x-0)(1-x)} \\ =\frac{2}{(1-x)(1-x)} \\ =\frac{2}{{(1-x)}^2} \end{gathered}$$

Hence,the value of$f\text{'}\left(x\right)=\frac{2}{{(1-x)}^2}$