- #1

- 208

- 7

## Main Question or Discussion Point

While working at home during the COVID-19 pandemic I've taken to seeing if I can still do math from undergrad (something I do once in a while to at least pretend my life isn't dominated by excel). So to that I've been reviewing partial derivatives (which I haven't really thought about in a good seven-ish years).

The exercise I'm working asks the following: Compute all first and second partial derivatives, including mixed derivatives, of the following ##0 = sin(xy) - x^2 -y^2##.

I think I'm good on the first and second partials:

$$\frac {\partial V}{\partial x} = y cos(xy) - 2x$$

$$\frac {\partial V}{\partial y} = x cos(xy) - 2y$$

$$\frac {\partial^2 V}{\partial x^2} = -y^2 sin(xy) - 2$$

$$\frac {\partial^2 V}{\partial y^2} = -x^2 sin(xy) - 2$$

Where I get tripped up is on the mixed derivative:

$$\frac {\partial^2 V}{\partial x \partial y} = \frac {\partial}{\partial x} \left( \frac {\partial V}{\partial y} \right) = \frac {\partial}{\partial x} \left( x cos(xy) - 2y \right)...$$

And this is where I get hung up. Do I simply proceed with an application of the production rule, ##(f \cdot g)' = f' \cdot g + f \cdot g'##?

$$\frac {\partial}{\partial x} \left( x cos(xy) - 2y \right) = (1) cos(xy) + x (-y sin(xy)) = cos(xy) - xy sin(xy)$$

I feel like I'm missing something. Have I done this right? Am I way overthinking this?

The exercise I'm working asks the following: Compute all first and second partial derivatives, including mixed derivatives, of the following ##0 = sin(xy) - x^2 -y^2##.

I think I'm good on the first and second partials:

$$\frac {\partial V}{\partial x} = y cos(xy) - 2x$$

$$\frac {\partial V}{\partial y} = x cos(xy) - 2y$$

$$\frac {\partial^2 V}{\partial x^2} = -y^2 sin(xy) - 2$$

$$\frac {\partial^2 V}{\partial y^2} = -x^2 sin(xy) - 2$$

Where I get tripped up is on the mixed derivative:

$$\frac {\partial^2 V}{\partial x \partial y} = \frac {\partial}{\partial x} \left( \frac {\partial V}{\partial y} \right) = \frac {\partial}{\partial x} \left( x cos(xy) - 2y \right)...$$

And this is where I get hung up. Do I simply proceed with an application of the production rule, ##(f \cdot g)' = f' \cdot g + f \cdot g'##?

$$\frac {\partial}{\partial x} \left( x cos(xy) - 2y \right) = (1) cos(xy) + x (-y sin(xy)) = cos(xy) - xy sin(xy)$$

I feel like I'm missing something. Have I done this right? Am I way overthinking this?