# What is the derivative of sqrt(2x)?

Dec 13, 2014

Power rule: $\frac{\mathrm{dy}}{\mathrm{dx}} \left[{x}^{n}\right] = n \cdot {x}^{n - 1}$

Power rule + chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} \left[{u}^{n}\right] = n \cdot {u}^{n - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = 2 x$ so $\frac{\mathrm{du}}{\mathrm{dx}} = 2$

We're left with $y = \sqrt{u}$ which can be rewritten as $y = {u}^{\frac{1}{2}}$

Now, $\frac{\mathrm{dy}}{\mathrm{dx}}$ can be found using the power rule and the chain rule.

Back to our problem: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cdot {u}^{- \frac{1}{2}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

plugging in $\frac{\mathrm{du}}{\mathrm{dx}}$ we get:

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

we know that: $\frac{2}{2} = 1$

therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = {u}^{- \frac{1}{2}}$

Plugging in the value for $u$ we find that:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{- \frac{1}{2}}$