Question

(7) 6. (a) give the limit definition of the derivative of a function $f(x)$
$f(x)=lim$

b) use the definition of derivative to find $f(x)$ for $f(x)=\frac{1}{x^{2}}$.

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$.

Step2: Substitute $f(x)=\frac{1}{x^{2}}$ into the definition

First, find $f(x + h)$: $f(x + h)=\frac{1}{(x + h)^{2}}$. Then, $\frac{f(x + h)-f(x)}{h}=\frac{\frac{1}{(x + h)^{2}}-\frac{1}{x^{2}}}{h}$.

Step3: Simplify the numerator

$\frac{1}{(x + h)^{2}}-\frac{1}{x^{2}}=\frac{x^{2}-(x + h)^{2}}{x^{2}(x + h)^{2}}=\frac{x^{2}-(x^{2}+2xh+h^{2})}{x^{2}(x + h)^{2}}=\frac{x^{2}-x^{2}-2xh - h^{2}}{x^{2}(x + h)^{2}}=\frac{-2xh - h^{2}}{x^{2}(x + h)^{2}}$.

Step4: Simplify the fraction $\frac{f(x + h)-f(x)}{h}$

$\frac{\frac{-2xh - h^{2}}{x^{2}(x + h)^{2}}}{h}=\frac{-2xh - h^{2}}{hx^{2}(x + h)^{2}}=\frac{h(-2x - h)}{hx^{2}(x + h)^{2}}=\frac{-2x - h}{x^{2}(x + h)^{2}}$.

Step5: Take the limit as $h

ightarrow0$
$f^{\prime}(x)=\lim_{h
ightarrow0}\frac{-2x - h}{x^{2}(x + h)^{2}}$. Substitute $h = 0$ into the expression: $f^{\prime}(x)=\frac{-2x-0}{x^{2}(x + 0)^{2}}=-\frac{2}{x^{3}}$.

Answer:

a) $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$
b) $f^{\prime}(x)=-\frac{2}{x^{3}}$