# How do you find the derivative of  -2- (1/(x^2)) + (4/(x^4))?

Jan 14, 2016

$f ' \left(x\right) = 2 \frac{{x}^{2} - 8}{x} ^ 5$

#### Explanation:

Remembering that:

$\frac{d}{\mathrm{dx}} {\sum}_{i = 1}^{n} {k}_{i} \cdot {f}_{i} \left(x\right) = \frac{d}{\mathrm{dx}} \left[{k}_{1} {f}_{1} \left(x\right) + {k}_{2} {f}_{2} \left(x\right) + \ldots + {k}_{n} {f}_{n} \left(x\right)\right] =$

$= {k}_{i} {\sum}_{i = 1}^{n} \frac{d}{\mathrm{dx}} {f}_{i} \left(x\right) = {k}_{1} {f}_{1} ' \left(x\right) + {k}_{2} {f}_{2} ' \left(x\right) + \ldots + {k}_{n} {f}_{n} ' \left(x\right)$

And

$\frac{d}{\mathrm{dx}} \left(g \frac{x}{h \left(x\right)}\right) = \frac{h ' \left(x\right) \cdot g \left(x\right) - g ' \left(x\right) \cdot h \left(x\right)}{{g}^{2} \left(x\right)}$

given:

$f \left(x\right) = - 2 - \frac{1}{x} ^ 2 + \frac{4}{x} ^ 4$

$\therefore f ' \left(x\right) = 0 - \left(\frac{0 \cdot {x}^{2} - 1 \cdot 2 x}{x} ^ 4\right) + 4 \cdot \left(\frac{0 \cdot {x}^{4} - 4 {x}^{3}}{x} ^ 8\right) =$

$= - \left(- 2 \frac{\cancel{x}}{{x}^{{\cancel{4}}^{3}}}\right) + 4 \cdot \left(- 4 \frac{\cancel{{x}^{3}}}{x} ^ \left({\cancel{8}}^{5}\right)\right) =$

$= \frac{2}{x} ^ 3 - \frac{16}{x} ^ 5 = \frac{2}{x} ^ 3 \left(1 - \frac{8}{x} ^ 2\right) = \frac{2}{x} ^ 3 \frac{{x}^{2} - 8}{x} ^ 2 = 2 \frac{{x}^{2} - 8}{x} ^ 5$