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Compute partial derivatives of the given function

by RoRi

Let

    \[ f(x,y) = x \sin (x+y). \]

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

From this we see that the mixed partials are equal:

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

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