How do you find the derivative of 2/sqrtx?

Jun 3, 2015

One law of exponentials states that ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$. Thus, we can rewrite the function as

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

Also, another law states that ${a}^{-} n = \frac{1}{a} ^ n$

Thus, we can rewrite this again as $2 {x}^{- \frac{1}{2}}$

Now, derivating the function:

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

If you prefer, $- \frac{1}{\sqrt{{x}^{3}}}$