Question

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=x^{2}-2 x y+3 y^{2} $$

Short answer

The second partial derivatives are \(\frac{\partial^2 z}{\partial x^2} = 2\), \(\frac{\partial^2 z}{\partial y^2} = 6\) and the mixed partial ones are \(\frac{\partial^2 z}{\partial y \partial x }= -2\), \(\frac{\partial^2 z}{\partial x \partial y }= -2\).

Step-by-step solution

Calculate the First Partial Derivatives

Taking the first partial derivative with respect to x: \[\frac{\partial z}{\partial x} = 2 x - 2 y\] And, the first partial derivative with respect to y: \[\frac{\partial z}{\partial y} = - 2 x + 6 y\]

Calculate the Second Partial Derivatives

Taking the second partial derivative with respect to x: \[\frac{\partial^2 z}{\partial x^2} = 2\] For y: \[\frac{\partial^2 z}{\partial y^2} = 6\] , And, the mixed second partial derivatives: \[\frac{\partial^2 z}{\partial y \partial x }= -2\ ,\ \frac{\partial^2 z}{\partial x \partial y }= -2\]

Confirm the equality of mixed partial derivatives

Check that the mixed second partial derivatives are equal: \[\frac{\partial^2 z}{\partial y \partial x } = \frac{\partial^2 z}{\partial x \partial y}\] Thus, they are indeed equal as they both equal to -2.