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Compute partial derivatives of the given function

by RoRi

Let

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

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} &= y + \frac{1}{y}; \\ \frac{\partial f}{\partial y} &= x - \frac{x}{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( y + \frac{1}{y} \right) \\ &= 0. \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} \left(y + \frac{1}{y} \right) \\ &= 1 - \frac{1}{y^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( x - \frac{x}{y^2} \right) \\ &= 1 - \frac{1}{y^2}.\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} \left( x - \frac{x}{y^2} \right) \\ &= \frac{2x}{y^3}. \end{align*}

From this we see that the mixed partials are equal:

    \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 1 - \frac{1}{y^2}. \]

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