# What is the derivative of (1-x^2)/(x^3)?

May 14, 2016

$\frac{{x}^{2} - 3}{{x}^{4}}$

#### Explanation:

Split up the fraction and rewrite it with negative exponents:

$\frac{1 - {x}^{2}}{x} ^ 3 = \frac{1}{x} ^ 3 - {x}^{2} / {x}^{3} = {x}^{-} 3 - {x}^{-} 1$

Now, differentiate ${x}^{-} 3 - {x}^{-} 1$ using the power rule, which states that the derivative of ${x}^{n} = n {x}^{n - 1}$. Thus:

$\frac{d}{\mathrm{dx}} \left({x}^{-} 3 - {x}^{-} 1\right) = \left(- 3\right) {x}^{- 3 - 1} - \left(- 1\right) {x}^{- 1 - 1}$

$= - 3 {x}^{-} 4 + {x}^{-} 2$

This is a valid final answer, but we can also write it without negative exponents by multiplying it by ${x}^{4} / {x}^{4}$:

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