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) = 1/x².

Explanation:

Step1: Recall derivative limit - def

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 formula

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

Step3: Simplify the numerator

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

Step4: Simplify the fraction

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

Step5: Find the limit as $h

ightarrow0$
$f^{\prime}(x)=\lim_{h
ightarrow0}\frac{-2x - h}{x^{2}(x + h)^{2}}$. Substituting $h = 0$ gives $f^{\prime}(x)=-\frac{2}{x^{3}}$.

Answer:

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