# How do you find the derivative of sqrt(4-x^2)?

May 23, 2015

Using chain rule, by naming $u = 4 - {x}^{2}$.

Before starting, let's just rewrite your fucntion

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

Now, as the chain rule states that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 {u}^{\frac{1}{2}}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$

Now

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \cancel{2} x}{\cancel{2} {u}^{\frac{1}{2}}}$

Substituting $u$:

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