Partial Derivative of f(x,y) = ∫xy cos(t2) dt?

[gnome]

I need the partial derivatives of:

f(x,y) = ∫_x^y cos(t²) dt

are they simply:

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

or am I completely lost here?

 

[NateTG]

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

?

 

[gnome]

No, I'm definitely not sure. Are you?

 

[NateTG]

Well, let's say that
∫cos(t²) dt = g(t) + C
then
dg/dt= cos(t²) from the fundamental theorem of calculus.

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

Does that make sense to you?

 

[gnome]

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[/color]]
and
∂f/∂y = df/dg * ∂g/∂y [edited: same reason as above[/color]]

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.

 

[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)

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(∫_x^ycos(t²)dt= (-1)cos(x²) and
∂/∂y(∫_x^ycos(t²)dt= (+1)cos(y²).

 

[NateTG]

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 fundamental theorem of calculus.

 

[NateTG]

gnome:

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

∫_x^yg'(t)dt=g(y)-g(x)

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

∫_x^ycos(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?

 

[gnome]

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.

 

[NateTG]

You've pretty much got it.

 

[HallsofIvy]

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.