Question

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=\sin (x-2 y) $$

Short answer

The four second partial derivatives of the function \( z = \sin(x - 2y) \) are \( f_{xx} = -\sin(x - 2y) \), \( f_{yy} = -4\sin(x - 2y) \), \( f_{xy} = 2\sin(x - 2y) \) and \( f_{yx} = 2\sin(x - 2y) \). The two mixed second partial derivatives are equal, verifying Clairaut's theorem.

Step-by-step solution

Find first partial derivatives

The first partial derivatives are calculated as follows: \[ f_x = \frac{\partial z}{\partial x} = \cos(x - 2y) \] and \[ f_y = \frac{\partial z}{\partial y} = -2\cos(x - 2y) \].

Find second partial derivatives

Now we will find the four second partial derivatives using the first partial derivatives. They are calculated as follows: \[ f_{xx} = \frac{\partial^2 z}{\partial x^2} = -\sin(x - 2y) \], \[ f_{yy} = \frac{\partial^2 z}{\partial y^2} = -4\sin(x - 2y) \], \[ f_{xy} = \frac{\partial^2 z}{\partial x \partial y} = 2\sin(x - 2y) \] and \[ f_{yx} = \frac{\partial^2 z}{\partial y \partial x} = 2\sin(x - 2y) \].

Confirm the equality of mixed partials

We can note that the second mixed partials \( f_{xy} \) and \( f_{yx} \) are equal, which verifies Clairaut's theorem, stating that for a function with continuous partial derivatives, the mixed partial derivatives are the same no matter the order of differentiation.