How do you find the derivative of $\sqrt{x^{2} + y^{2}}$ $sqrt(x^2+y^2)$ ?

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How do you find the derivative of $\sqrt{x^{2} + y^{2}}$ $sqrt(x^2+y^2)$ ?

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Answer 1 · 2015-06-23T00:34:32

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

Explanation:

$\frac{d}{d x} (\sqrt{x}) = \frac{1}{2 \sqrt{x}}$ $d/dx(sqrtx) = 1/(2sqrtx)$ , so

$\frac{d}{d x} (\sqrt{u}) = \frac{1}{2 \sqrt{u}} \frac{d u}{d x}$ $d/dx(sqrtu) = 1/(2sqrtu) (du)/dx$ .

$\frac{d}{d x} (\sqrt{x^{2} + y^{2}}) = \frac{1}{2 \sqrt{x^{2} + y^{2}}} \cdot \frac{d}{d x} (x^{2} + y^{2})$ $d/dx(sqrt(x^2+y^2)) = 1/(2sqrt(x^2+y^2)) * d/dx(x^2+y^2)$

$= \frac{1}{2 \sqrt{x^{2} + y^{2}}} (2 x + 2 y \frac{d y}{d x})$ $= 1/(2sqrt(x^2+y^2))(2x+2y dy/dx)$

$= \frac{1}{2 \sqrt{x^{2} + y^{2}}} 2 x + \frac{1}{2 \sqrt{x^{2} + y^{2}}} 2 y \frac{d y}{d x}$ $=1/(2sqrt(x^2+y^2))2x+ 1/(2sqrt(x^2+y^2))2y dy/dx$

$= \frac{x}{\sqrt{x^{2} + y^{2}}} + \frac{y}{\sqrt{x^{2} + y^{2}}} \frac{d y}{d x}$ $=x/sqrt(x^2+y^2)+ y/sqrt(x^2+y^2) dy/dx$

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

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How do you find the derivative of $\sqrt{x^{2} + y^{2}}$ $sqrt(x^2+y^2)$ ?

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How do you find the derivative of $\sqrt{x^{2} + y^{2}}$ $sqrt(x^2+y^2)$ ?