Question

Use the definition of the derivative to find $f\text{'}$for each function $f$in Exercises 34-59

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

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

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

Putting there values in (1)

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

Putting $h=0$

$$\begin{gathered} =\frac{-{\left(0\right)}^2-3x^2-3x\left(0\right)}{x^3{(x+0)}^3} \\ =\frac{0-3x^2-0}{x^3\left(x^3\right)} \\ =\frac{-3x^2}{x^6} \\ =\frac{-3}{x^4} \end{gathered}$$

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