# How do you find the derivative of sqrt(1/x^3)?

Mar 15, 2016

Rewrite it. $\sqrt{\frac{1}{{x}^{3}}} = {x}^{- \frac{3}{2}}$. So the derivative is $- \frac{3}{2} {x}^{- \frac{5}{2}} = \frac{- 3}{2 \sqrt{{x}^{5}}}$

#### Explanation:

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

Now apply the power rule.

If you really like the quotient rule and algebra , then use

$\sqrt{\frac{1}{{x}^{3}}} = \frac{1}{\sqrt{{x}^{3}}} = \frac{1}{x} ^ \left(\frac{3}{2}\right)$ and the quotient rule.

So the derivative is $\frac{\left(0\right) {x}^{\frac{3}{2}} - 1 \left(\frac{3}{2} {x}^{\frac{1}{2}}\right)}{{x}^{\frac{3}{2}}} ^ 2$.

Now simplify algebraically to get $\frac{- 3}{2 \sqrt{{x}^{5}}}$.

If you are a glutton for algebra use the chain rule and the quotient rule:

$\frac{d}{\mathrm{dx}} \left(\sqrt{\frac{1}{x} ^ 3}\right) = \frac{d}{\mathrm{dx}} \left({\left(\frac{1}{x} ^ 3\right)}^{\frac{1}{2}}\right)$

$= \frac{1}{2} {\left(\frac{1}{x} ^ 3\right)}^{- \frac{1}{2}} \left[\frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 3\right)\right]$

$= \frac{1}{2} {\left(\frac{1}{x} ^ 3\right)}^{- \frac{1}{2}} \left[\frac{\left(0\right) {x}^{3} - 1 \left(3 {x}^{2}\right)}{{x}^{3}} ^ 2\right]$

Now simplify algebraically.