Understanding Mixed Partial Derivatives: How Do You Solve Them?

[atomicpedals]

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?

 

[PeroK]

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?

That looks right.

 

[atomicpedals]

That looks right.

Thanks! Sometimes, when you spend too long looking at an exercise, you question your sanity.

 

[HallsofIvy]

To take a partial derivative with respect to x you treat y as a constant. So $\frac{\partial}{\partial x}x cos(xy)- 2x$ is the same as $\frac{d}{dx} xcos(ax)- 2x= cos(ax)- axsin(x)- 2$ where, since I replaced "y" with the constant "a" we need to replace back- $\frac{\partial^2V}{\partial xy}= cos(xy)- xysin(xy)- 2$.

Similarly, $\frac{\partial}{\partial y} y cox(xy)- 2y$ is $\frac{d}{dy} y cos(ay)- 2y= cos(ay)- ay sin(ay)- 2= cos(xy)- xy sin(xy)- 2$.

Notice that those are the same. It is true that, as long as V and its first and second partial derivatives are continuous, $\frac{\partial}{\partial x}\frac{\partial V}{\partial y}= \frac{\partial}{\partial y}\frac{\partial V}{\partial x}$.