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

In summary: As I stated above, all you need to do is apply the fundamental theorem of calculus. Which is exactly what I did.In summary, the conversation discusses the partial derivatives of a function f(x,y) that involves an integral of xy cos(t2) dt. The conversation includes various methods of finding the partial derivatives, such as using the fundamental theorem of calculus and Leibniz's formula. There is also a discussion about the use of variables in the integral and how to handle functions of x in the limits of integration. Ultimately, it is concluded that the partial derivatives are ∂f/∂x = -2xcos(x2) and ∂f/∂y = 2ycos(y2).
  • #1
gnome
1,041
1
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?
 
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  • #2
Are you sure it's not
∂f/∂x=-cos(x2)
and
∂f/∂y=-cos(y2)

?
 
  • #3
No, I'm definitely not sure. Are you?
 
  • #4
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
∫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
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.
 
Last edited:
  • #6
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
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.
 
  • #8
gnome:

Let's digress for a moment and look at the fundamental 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
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
You've pretty much got it.
 
  • #11
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.
 

1. What is a partial derivative?

A partial derivative is a mathematical concept that measures how much a function changes when one of its variables is changed, while holding all other variables constant.

2. How is a partial derivative calculated?

A partial derivative is calculated by finding the derivative of the function with respect to a specific variable, while treating all other variables as constants. This can be done using the standard rules of differentiation.

3. What is the significance of the notation ∫ in the function?

The notation ∫ represents an indefinite integral, which is the inverse operation of differentiation. In this case, it indicates that the function is the antiderivative of the integrand (xy cos(t2)).

4. Can the partial derivative of a function change?

Yes, the partial derivative of a function can change depending on which variable is being held constant. This is because the function may have different rates of change with respect to different variables.

5. How is the partial derivative related to the total derivative?

The partial derivative is one component of the total derivative, which measures the overall rate of change of a function. The total derivative takes into account the effects of all variables, while the partial derivative only considers the change in one variable.

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