- #1

- 1,036

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f(x,y) = ∫

_{x}

^{y}cos(t

^{2}) dt

are they simply:

∂f/∂x = -2xcos(x

^{2})

and

∂f/∂y = 2ycos(y

^{2})

or am I completely lost here?

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- #1

- 1,036

- 1

f(x,y) = ∫

are they simply:

∂f/∂x = -2xcos(x

and

∂f/∂y = 2ycos(y

or am I completely lost here?

- #2

NateTG

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Are you sure it's not

∂f/∂x=-cos(x^{2})

and

∂f/∂y=-cos(y^{2})

?

∂f/∂x=-cos(x

and

∂f/∂y=-cos(y

?

- #3

- 1,036

- 1

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

- #4

NateTG

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∫cos(t

then

dg/dt= cos(t

Now we have

∫

so I get

∂f/∂x=-cos(x

∂f/∂y=cos(y

Does that make sense to you?

- #5

- 1,036

- 1

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) = t^{2} 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(y^{2}) - cos(x^{2}) 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(y^{2}) - cos(x^{2})] = -2xcos(x^{2})

and

∂f/∂y = ∂/∂y[cos(y^{2}) - cos(x^{2})] = 2ycos(y^{2})

But I'm not sure I have the f's and g's right, & maybe all of that is nonsense.

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

I'm thinking that we have g(t) = t

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(y

and

∂f/∂y = ∂/∂y[cos(y

But I'm not sure I have the f's and g's right, & maybe all of that is nonsense.

Last edited:

- #6

HallsofIvy

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d/dx(∫

(db/dx)f(x,b(x))- (da/dx)f(x,a(x)+ ∫

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(∫

∂/∂y(∫

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

∫

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

∫

Now, since g'(t)=cos(t

∂/∂x g(x) = g'(x)=cos(x

and

∂/∂x g(y) = 0

Does that make sense?

- #9

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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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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 think you're mixing your variables. You should probably use something other than f and x 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.I also don't understand how Leibnitz's formula applies, since there is no multi-variable function inside the integral.

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