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Compute partial derivatives of the given function

by RoRi

Let

    \[ f(x,y) = \sqrt{x^2 + y^2}. \]

Compute all first-order and second-order partial derivatives. Verify that the mixed partial derivatives are equal.

The first-order partial derivatives are given by

    \begin{align*} \frac{\partial f}{\partial x} &= \left( \frac{1}{2} \right) \left( \frac{1}{\sqrt{x^2+y^2}} \right) (2x) = \frac{x}{\sqrt{x^2+y^2}}; \\ \frac{\partial f}{\partial y} &= \left( \frac{1}{2} \right) \left( \frac{1}{\sqrt{x^2+y^2}} \right) (2y) = \frac{y}{\sqrt{x^2+y^2}}. \end{align*}

The second partial derivatives are given by

    \begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2+y^2}} \right) \\ &= \frac{\sqrt{x^2+y^2} - \frac{x^2}{\sqrt{x^2+y^2}}}{x^2+y^2} \\ &= \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}. \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} \left( \frac{x}{\sqrt{x^2+y^2}} \right) \\ &= \frac{-xy}{(x^2+y^2)^{\frac{3}{2}}}.\\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} \left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{-xy}{(x^2+y^2)^{\frac{3}{2}}}.\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} \left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{x^2}{(x^2+y^2)^{\frac{3}{2}}}. \end{align*}

From this we see that the mixed partials are equal:

    \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = \frac{-xy}{(x^2+y^2)^{\frac{3}{2}}}. \]

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