Partial Derivative

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  • #1
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Main Question or Discussion Point

I need the partial derivatives of:

f(x,y) = ∫xy cos(t2) dt

are they simply:

∂f/∂x = -2xcos(x2)
and
∂f/∂y = 2ycos(y2)

or am I completely lost here?
 

Answers and Replies

  • #2
NateTG
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Are you sure it's not
∂f/∂x=-cos(x2)
and
∂f/∂y=-cos(y2)

?
 
  • #3
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No, I'm definitely not sure. Are you?
 
  • #4
NateTG
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Well, lets say that
∫cos(t2) dt = g(t) + C
then
dg/dt= cos(t2) from the fundemental theorem of calculus.

Now we have
∫xycos(t2) dt=g(y)-g(x)
so I get
∂f/∂x=-cos(x2)
∂f/∂y=cos(y2)

Does that make sense to you?
 
  • #5
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I'm not sure. It's confusing.

Isn't this a composite function that calls for use of the chain rule?

I'm thinking that we have g(t) = t2 and the integral is f(g(t)) which gets evaluated at t=y and t=x, thus becoming f(g(x)) and f(g(y)), so cos(y2) - cos(x2) is df/dg and we still have to differentiate that wrt x to get ∂f/∂x and wrt y to get ∂f/∂y.

So, I guess I'm saying
∂f/∂x = df/dg * ∂g/∂x [edited: I had the derivatives & the partials reversed]
and
∂f/∂y = df/dg * ∂g/∂y [edited: same reason as above]

Then,
∂f/∂x = ∂/∂x[cos(y2) - cos(x2)] = -2xcos(x2)
and
∂f/∂y = ∂/∂y[cos(y2) - cos(x2)] = 2ycos(y2)

But I'm not sure I have the f's and g's right, & maybe all of that is nonsense.
 
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  • #6
HallsofIvy
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Leibniz's formula is very general:

d/dx(∫b(x)a(x)f(x,t)dt)=

(db/dx)f(x,b(x))- (da/dx)f(x,a(x)+ ∫b(x)a(x)(∂f(x,t)/∂t dt)

It doesn't matter that you are dealing with two variables, x, y, since with partial derivatives you are treating one of them as a constant.

In particular, ∂/∂x(∫xycos(t2)dt= (-1)cos(x2) and
∂/∂y(∫xycos(t2)dt= (+1)cos(y2).
 
  • #7
NateTG
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Originally posted by HallsofIvy
Leibniz's formula is very general:

d/dx(∫b(x)a(x)f(x,t)dt)=

(db/dx)f(x,b(x))- (da/dx)f(x,a(x)+ ∫b(x)a(x)(∂f(x,t)/∂t dt)
I think you're mixing your variables. You should probably use something other than f and x inside the integral.

I also don't understand how Leibnitz's formula applies, since there is no multi-variable function inside the integral.

As I stated above, all you need to do is apply the fundemental theorem of calculus.
 
  • #8
NateTG
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gnome:

Let's digress for a moment and look at the fundemental theorem of calculus:

∫xyg'(t)dt=g(y)-g(x)

Now, let's say we have some g(t) so that g'(t)=cos(t2) so:

∫xycos(t2)dt=g(y)-g(x)

Now, since g'(t)=cos(t2) we get
∂/∂x g(x) = g'(x)=cos(x2)
and
∂/∂x g(y) = 0

Does that make sense?
 
  • #9
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Well, I'm sure you're right & I'm sitting here with my textbook from calc I opened to the Fundamental Theorem of Calculus trying to make sense of it. To simplify matters, I'm going to forget about partial derivatives for the moment, & just look at these:

f(x) = cos(2x-1)
f'(x) = -2sin(2x-1) (of this, I'm certain)

-------------------------------------

Now, what you're telling me is

g(x) = ∫sin(2x-1)dx
g'(x) = sin(2x-1)

Right?

And the more I think about that, the more sense it seems to make. But if you can add anything to that to make it clearer, please do.

Thanks.
 
  • #10
NateTG
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You've pretty much got it.
 
  • #11
HallsofIvy
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I think you're mixing your variables. You should probably use something other than f and x inside the integral.
No, I was not mixing variables. It was necessary to use f and x "inside the integral" because they appeared outside the intergral. The only dummy variable was t.

I also don't understand how Leibnitz's formula applies, since there is no multi-variable function inside the integral.
Leibnitz's formula still applies, the last integral happens to be 0. The original question was about how you handled functions of x in the limits of integration. Leibnitz's formula does that nicely.
 

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