What is the derivative of $\sqrt{\frac{1}{x^{3}}}$ $sqrt(1/x^3)$ ?

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What is the derivative of $\sqrt{\frac{1}{x^{3}}}$ $sqrt(1/x^3)$ ?

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Answer 1 · 2016-04-25T23:57:38

$- \frac{3}{2} x^{- \frac{5}{2}}$ $-3/2x^(-5/2)$

Explanation:

The most important thing here is not calculus but algebra. In particular, the properties of exponents.

Note that $\sqrt{\frac{1}{x^{3}}}$ $sqrt(1/x^3)$ is equivalent to $\sqrt{x^{- 3}}$ $sqrt(x^(-3))$ (because $\frac{1}{a}$ $1/a$ is equal to $a^{- 1}$ $a^(-1)$ ). Using the property $\sqrt[a]{x^{b}} = x^{\frac{b}{a}}$ $root(a)(x^b)=x^(b/a)$ , $\sqrt[2]{x^{- 3}} = x^{- \frac{3}{2}}$ $root(2)(x^(-3))=x^(-3/2)$ . Our problem now is simply finding the derivative of $x^{- \frac{3}{2}}$ $x^(-3/2)$ , which is done easily using the power rule:
$\frac{d}{d x} x^{- \frac{3}{2}} = - \frac{3}{2} \cdot x^{- \frac{3}{2} - 1} = - \frac{3}{2} x^{- \frac{5}{2}}$ $d/dxx^(-3/2)=-3/2*x^(-3/2-1)=-3/2x^(-5/2)$

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What is the derivative of $\sqrt{\frac{1}{x^{3}}}$ $sqrt(1/x^3)$ ?

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