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Compute partial derivatives of the given function

by RoRi

Let

    \[ f(x,y) = \frac{x}{\sqrt{x^2+y^2}} \qquad (x,y) \neq (0,0). \]

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} &= \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 f}{\partial y} &= \frac{-x \frac{y}{\sqrt{x^2+y^2}}}{x^2+y^2} \\ &= \frac{-xy}{(x^2+y^2)^{\frac{3}{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{y^2}{(x^2+y^2)^{\frac{3}{2}}} \right) \\ &= \frac{-y^2 (3/2)(2x)(\sqrt{x^2+y^2})}{(x^2+y^2)^3} \\ &= \frac{-3xy^2}{(x^2+y^2)^{\frac{5}{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{y^2}{(x^2+y^2)^{\frac{3}{2}}} \right) \\ &= \frac{2y(x^2+y^2)^{\frac{3}{2}} - y^2 (3/2)(2y)\sqrt{x^2+y^2}}{(x^2+y^2)^3} \\ &= \frac{2y(x^2+y^2) - 3y^3}{(x^2+y^2)^{\frac{5}{2}}} \\ &= \frac{y(2x^2 - y^2)}{(x^2+y^2)^{\frac{5}{2}}}. \end{align*}

Then,

    \begin{align*} \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{-xy}{(x^2+y^2)^{\frac{3}{2}}} \right) \\ &= \frac{-y(x^2+y^2)^{\frac{3}{2}} + xy(2x)(3/2)\sqrt{x^2+y^2}}{(x^2+y^2)^3} \\ &= \frac{-yx^2 - y^3 + 3x^2y}{(x^2+y^2)^{\frac{5}{2}}} \\ &= \frac{y (2x^2 y - y^2)}{(x^2+y^2)^{\frac{5}{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{-xy}{(x^2+y^2)^{\frac{3}{2}}}\right) \\ &= \frac{-x(x^2+y^2)^{\frac{3}{2}} + xy(2y)(3/2) \sqrt{x^2+y^2}}{(x^2+y^2)^3}\\ &= \frac{-x^3 -xy^2 + 3xy^2}{(x^2+y^2)^{\frac{5}{2}}} \\ &= \frac{x (2xy^2 - x^2)}{(x^2+y^2)^{\frac{5}{2}}}. \end{align*}

From this we see that the mixed partials are equal:

    \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = \frac{y (2x^2 y - y^2)}{(x^2+y^2)^{\frac{5}{2}}}. \]

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