# How do you find f'(x) using the definition of a derivative for f(x)=(4/x^2) ?

Oct 20, 2015

See the explanation.

#### Explanation:

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

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{4}{x + h} ^ 2 - \frac{4}{x} ^ 2}{h} = {\lim}_{h \to 0} \frac{\frac{4 {x}^{2} - 4 {\left(x + h\right)}^{2}}{{x}^{2} {\left(x + h\right)}^{2}}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{4 {x}^{2} - 4 {x}^{2} - 8 x h - 4 {h}^{2}}{h {x}^{2} {\left(x + h\right)}^{2}}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{- 8 x h - 4 {h}^{2}}{h {x}^{2} {\left(x + h\right)}^{2}} = {\lim}_{h \to 0} \frac{- 8 x - 4 h}{{x}^{2} {\left(x + h\right)}^{2}} = \frac{- 8 x}{{x}^{2} {x}^{2}}$

$f ' \left(x\right) = - \frac{8}{x} ^ 3$