# How do you find the derivative of sqrt(x^2+y^2)?

Jun 23, 2015

To find $\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + {y}^{2}}\right)$, as part of an implicit differentiation problem, use the chain rule.

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$, so

$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \frac{\mathrm{du}}{\mathrm{dx}}$.

$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + {y}^{2}}\right) = \frac{1}{2 \sqrt{{x}^{2} + {y}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2}\right)$

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

$= \frac{1}{2 \sqrt{{x}^{2} + {y}^{2}}} 2 x + \frac{1}{2 \sqrt{{x}^{2} + {y}^{2}}} 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

$= \frac{x}{\sqrt{{x}^{2} + {y}^{2}}} + \frac{y}{\sqrt{{x}^{2} + {y}^{2}}} \frac{\mathrm{dy}}{\mathrm{dx}}$

In order to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ you will, of course, need the rest of the derivative of the rest of the original equation.