Question

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=x^{3}-4 y^{2} $$

Short answer

The second partial derivatives are: \( \frac{\partial^2 z}{\partial x^2} = 6x \), \( \frac{\partial^2 z}{\partial y^2} = -8 \), \( \frac{\partial^2 z}{\partial x \partial y} = 0 \), and \( \frac{\partial^2 z}{\partial y \partial x} = 0 \). The mixed partials are equal, as suggested by Clairaut's theorem.

Step-by-step solution

Compute the first partial derivatives

The first partial derivative with respect to x is given by \( \frac{\partial z}{\partial x} = 3x^{2} \). Similarly, the first partial derivative with respect to y is \( \frac{\partial z}{\partial y} = -8y \).

Compute the second partial derivatives

The second partial derivative with respect to x is given by \( \frac{\partial^2 z}{\partial x^2} = 6x \). Similarly, the second partial derivative with respect to y is \( \frac{\partial^2 z}{\partial y^2} = -8 \).

Compute mixed partial derivatives

For the second mixed partial derivative with respect to x and then y, differentiating \( \frac{\partial z}{\partial x} = 3x^{2} \) with respect to y gives \( \frac{\partial^2 z}{\partial x \partial y} = 0 \). Similarly, differentiating \( \frac{\partial z}{\partial y} = -8y \) with respect to x gives \( \frac{\partial^2 z}{\partial y \partial x} = 0 \). As expected from Clairaut's theorem, the mixed partials are indeed equal.