Question

find the derivative of the function using the definition of derivative.

$f(x)=\frac{1}{x^{2}-9}$

$f(x)=square$

state the domain of the function. (enter your answer using interval notation.)

$square$

state the domain of its derivative. (enter your answer using interval notation.)

$square$

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function $y = f(x)$ is $f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\frac{1}{x^{2}-9}$, then $f(x + h)=\frac{1}{(x + h)^{2}-9}=\frac{1}{x^{2}+2xh+h^{2}-9}$.

Step2: Calculate $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=\frac{1}{x^{2}+2xh + h^{2}-9}-\frac{1}{x^{2}-9}\\ &=\frac{(x^{2}-9)-(x^{2}+2xh+h^{2}-9)}{(x^{2}+2xh+h^{2}-9)(x^{2}-9)}\\ &=\frac{x^{2}-9 - x^{2}-2xh - h^{2}+9}{(x^{2}+2xh+h^{2}-9)(x^{2}-9)}\\ &=\frac{-2xh - h^{2}}{(x^{2}+2xh+h^{2}-9)(x^{2}-9)} \end{align*}$$

\]

Step3: Calculate $\frac{f(x + h)-f(x)}{h}$

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{\frac{-2xh - h^{2}}{(x^{2}+2xh+h^{2}-9)(x^{2}-9)}}{h}\\ &=\frac{-2x - h}{(x^{2}+2xh+h^{2}-9)(x^{2}-9)} \end{align*}$$

\]

Step4: Find the limit as $h

ightarrow0$
\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{-2x - h}{(x^{2}+2xh+h^{2}-9)(x^{2}-9)}\\ &=\frac{-2x}{(x^{2}-9)^{2}} \end{align*}$$

\]

Step5: Find the domain of the function $f(x)$

The function $f(x)=\frac{1}{x^{2}-9}$ is undefined when $x^{2}-9 = 0$, i.e., when $x=- 3$ or $x = 3$. So the domain of $f(x)$ is $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$.

Step6: Find the domain of the derivative $f'(x)$

The derivative $f'(x)=\frac{-2x}{(x^{2}-9)^{2}}$ is undefined when $(x^{2}-9)^{2}=0$, i.e., when $x=-3$ or $x = 3$. So the domain of $f'(x)$ is $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$.

Answer:

$f'(x)=\frac{-2x}{(x^{2}-9)^{2}}$
$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$
$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$