# What is the derivative of 1/sqrtx?

Jun 2, 2015

Oe law of exponentials states that ${a}^{-} n = \frac{1}{a} ^ n$. Also, another law states that ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$. Finally, we'll need a third one: ${\left({a}^{n}\right)}^{m} = {a}^{n \cdot m}$

Thus, we can rewrite this function following both laws. Then, we derivate it.

First law applied:

${\left(\sqrt{x}\right)}^{-} 1$

Second law:

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

Third law mentioned:

${x}^{- \frac{1}{2}}$

Now, deriving $f \left(x\right) = {x}^{- \frac{1}{2}}$

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

Or, if you prefer the root: $- \frac{1}{2 \sqrt{{x}^{3}}}$