# What is the derivative of 4/sqrtx?

Jun 2, 2015

Following an exponential rule that states ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$, we can rewrite the fucntion as

$\frac{4}{x} ^ \left(\frac{1}{2}\right)$

Another exponential rule states that $\frac{1}{a} ^ n = {a}^{-} n$

Thus,

$\frac{4}{x} ^ \left(\frac{1}{2}\right) = 4 {x}^{-} \left(\frac{1}{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- \frac{1}{2}\right) 4 {x}^{- \frac{3}{2}} = - 2 {x}^{- \frac{3}{2}}$

Or, if you prefer the original form, we can go back to the fraction and root:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{\sqrt{{x}^{3}}}$