Partial Derivative Question

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  • Thread starter atomicpedals
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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?
 

Answers and Replies

  • #2
PeroK
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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.
 
  • #3
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That looks right.
Thanks! Sometimes, when you spend too long looking at an exercise, you question your sanity.
 
  • #4
HallsofIvy
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To take a partial derivative with respect to x you treat y as a constant. So [itex]\frac{\partial}{\partial x}x cos(xy)- 2x[/itex] is the same as [itex]\frac{d}{dx} xcos(ax)- 2x= cos(ax)- axsin(x)- 2[/itex] where, since I replaced "y" with the constant "a" we need to replace back- [itex]\frac{\partial^2V}{\partial xy}= cos(xy)- xysin(xy)- 2[/itex].

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

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

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