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Compute partial derivatives of the given function

by RoRi

Let

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

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{1}{x-y} - \frac{x+y}{(x-y)^2} = \frac{-2y}{(x-y)^2}; \\ \frac{\partial f}{\partial y} &= \frac{1}{x-y} + \frac{x+y}{(x-y)^2} = \frac{2x}{(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( \frac{-2y}{(x-y)^2} \right) \\ &= \frac{(-2y)(2)(x-y)}{(x-y)^4} \\ &= \frac{4y}{(x-y)^3}. \\ \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{-2y}{(x-y)^2} \right) \\ &= \frac{2(x-y)^2 + 2y(2)(x-y)}{(x-y)^4}\\ &= \frac{-2 (x+y)}{(x-y)^3}. \\ \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{2x}{(x-y)^2} \right) \\ &= \frac{2(x-y)^2 - (2x)(2)(x-y)}{(x-y)^4} \\ &= \frac{-2(x+y)}{(x-y)^3}.\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} \left(\frac{2x}{(x-y)^2}\right) \\ &= \frac{(2x)(2)(x-y)}{(x-y)^4} \\ &= \frac{4x}{(x-y)^3}. \end{align*}

From this we see that the mixed partials are equal:

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

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