# How do you find the derivative of ((x^3)-(7/x))^-2 ?

Feb 12, 2017

$\frac{- 2 x \left(3 {x}^{4} + 7\right)}{{x}^{4} - 7} ^ 3$

#### Explanation:

Rewrite & then use a combination of the power rule & chain rule :
$y = {\left({x}^{3} - 7 {x}^{-} 1\right)}^{-} 2$

$y ' = - 2 {\left({x}^{3} - 7 {x}^{-} 1\right)}^{-} 3 \left(3 {x}^{2} + 7 {x}^{-} 2\right)$

Simplify by making all exponents positive & find common denominators:

$y ' = \frac{- 2 \left(3 {x}^{2} + \frac{7}{x} ^ 2\right)}{{x}^{3} - \frac{7}{x}} ^ 3 = \frac{- 2 \left(\frac{3 {x}^{4} + 7}{x} ^ 2\right)}{\frac{{x}^{4} - 7}{x}} ^ 3$

Simplify by cubing both numerator & denominator of the denominator term and multiplying reciprocals:
$y ' = - 2 \left(\frac{3 {x}^{4} + 7}{x} ^ 2\right) \frac{{x}^{3}}{{x}^{4} - 7} ^ 3$

Simplify and rearrange to get the final answer:
$\frac{- 2 x \left(3 {x}^{4} + 7\right)}{{x}^{4} - 7} ^ 3$