# Using the limit definition, how do you find the derivative of f(x)=3(x^(-2)) ?

Feb 11, 2016

$f ' \left(x\right) = - 6 {x}^{- 3}$

#### Explanation:

According to the limit definition, the derivative of $f \left(x\right)$ is

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{3 {\left(x + h\right)}^{- 2} - 3 {x}^{- 2}}{h}$

$= {\lim}_{h \to 0} \frac{\frac{3}{x + h} ^ 2 - \frac{3}{x} ^ 2}{h}$

$= {\lim}_{h \to 0} \frac{\frac{3 {x}^{2} - 3 {\left(x + h\right)}^{2}}{{\left(x + h\right)}^{2} \cdot {x}^{2}}}{h}$

$= {\lim}_{h \to 0} \frac{3 {x}^{2} - 3 {\left(x + h\right)}^{2}}{h {\left(x + h\right)}^{2} \cdot {x}^{2}}$

... use ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$...

$= {\lim}_{h \to 0} \frac{3 {x}^{2} - 3 \left({x}^{2} + 2 x h + {h}^{2}\right)}{h {\left(x + h\right)}^{2} \cdot {x}^{2}}$

$= {\lim}_{h \to 0} \frac{\cancel{3 {x}^{2}} - \cancel{3 {x}^{2}} - 6 x h - 3 {h}^{2}}{h {\left(x + h\right)}^{2} \cdot {x}^{2}}$

$= {\lim}_{h \to 0} \frac{h \left(- 6 x - 3 h\right)}{h {\left(x + h\right)}^{2} \cdot {x}^{2}}$

... cancel $h$....

$= {\lim}_{h \to 0} \frac{- 6 x - 3 h}{{\left(x + h\right)}^{2} \cdot {x}^{2}}$

... at this point you can apply the $\lim$, so plug $h = 0$:

$= \frac{- 6 x - 0}{{\left(x + 0\right)}^{2} \cdot {x}^{2}}$

$= \frac{- 6 x}{{x}^{4}}$

$= \frac{- 6}{{x}^{3}}$

$= - 6 {x}^{- 3}$

$f ' \left(x\right) = - 6 {x}^{- 3}$