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Compute partial derivatives of the given function

by RoRi

Let

    \[ f(x,y) = \sin (\cos (2x-3y)). \]

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

Using the chain rule, the first-order partial derivatives are given by

    \begin{align*} \frac{\partial f}{\partial x} &= -2 \cos (\cos (2x-3y)) \sin (2x-3y); \\ \frac{\partial f}{\partial y} &= 3 \cos (\cos (2x-3y)) \sin (2x-3y). \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} (-2 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= -4(\cos(2x-3y) \cos (\cos (2x-3y)) + (\sin (2x-3y))^2 \sin (\cos (2x-3y))). \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} (-2 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= 6\cos (2x-3y) \cos (\cos (2x-3y)) + 6 (\sin (2x-3y))^2 \sin (\cos (2x-3y)).\\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} (3 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= 6 \cos (2x-3y) \cos (\cos (2x-3y)) + 6(\sin (2x-3y))^2 \sin (\cos (2x-3y)).\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} (3 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= -9(\cos(2x-3y)\cos(\cos (2x-3y)) + (\sin (2x-3y))^2 \sin (\cos (2x-3y))). \end{align*}

From this we see that the mixed partials are equal:

    \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 6 \cos (2x-3y) \cos (\cos (2x-3y) + 6(\sin (2x-3y))^2 \sin (\cos (2x-3y)). \]

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