Partial Derivatives of an Integral

[knowLittle]

Homework Statement

Find the partial derivatives:
f(x,y)= integral[x,y] cos(t^2)dt, find f_x(x,y) and f_y(x,y)

Homework Equations

I know from calculus that the derivative of an integral is the function.

The Attempt at a Solution

I found that the integral of [x to y] cos(t^2)dt=sin(t^2)= sin(y^2)-sin(x^2)

and then found the partial derivative of this.
fx=-2xcos(x^2)
fy=2ycos(y^2)

I have seen other procedures and I think I'm wrong. Could someone help?

 

[Fredrik]

I found that the integral of [x to y] cos(t^2)dt=sin(t^2)

This is wrong. One way to see that is to compute d/dt sin(t^2).

You need to use the idea you described under "relevant equations". You should start by making your statement of that idea more precise. If g:ℝ→ℝ, then d/dx (what?) =g(x).

 

[knowLittle]

differentiate the integral as a function of x, how is that possible?
I would need a relation between x or y and t; however, none is shown other than the integral and the function.

 

[Fredrik]

If you can fill in the "what?" in the formula I typed, the rest is easy. t can't have a relationship with either x or y, since t is being integrated over.

 

[knowLittle]

I don't know what to write instead of "what". It could be the integral [x y]: cos(x^2)dx

 

[DrewD]

Be more general. Fill in the blank with a general expression and then apply it to this problem. How can one, in general, express the antiderivative of $g(x)$? You already pointed this out in your first post. Now write the general expression.

 

[Fredrik]

I don't know what to write instead of "what". It could be the integral [x y]: cos(x^2)dx

Forget about this problem for now, and just explain what you meant when you said that you know that "the derivative of an integral is the function"?

 

[knowLittle]

Ok.
I have: integral of [ f(x) ], then the derivative of: integral of [f(x)] = f(x).

or (d/dx)(integral(f(x))=f(x)

Thank you for your help so far. I can't see things clearly often.

 

[Fredrik]

You're still not specific enough. What are the limits of the integration in the formula d/dx[some kind of integral involving f]=f(x)?

 

[knowLittle]

(d/dx)INTEGRAL{f(x)dx }=f(x)

 

[Fredrik]

No, you need to specify what interval you're integrating over.

Edit: OK, I'll just tell you: For all $a\in\mathbb R$,
$$\frac{d}{dx}\int_a^x f(t)dt=f(x).$$

 

[knowLittle]

If you can fill in the "what?" in the formula I typed, the rest is easy. t can't have a relationship with either x or y, since t is being integrated over.

I thought that "t" can't have a relationship with either x or y?
Why do you relate them?

 

[Fredrik]

I didn't. t clearly doesn't depend on x, and since it makes no difference what variable you use as the dummy variable in the integration (i.e. you can substitute s for t without changing the value of the integral), x doesn't depend on t either.

 

[knowLittle]

Ok, so "t" can be any value from [a , x]?

 

[Fredrik]

I wouldn't say that. Consider the simpler example $\sum_{k=1}^2a_k$. What value does k have here? The notation is just an abbreviation for $a_1+a_2$, and k doesn't even appear in that. k only appears in the abbreviation, and can be replaced with any other variable without changing the meaning of the notation. So I wouldn't say that k can be any value in {1,2}. I would say that k isn't even a part of the expression we're working with, and that it's never assigned a value.

Similarly, $\int_a^x f(t)dt$ is an abbreviation for something that doesn't involve t. This is why these things are called "dummy variables".

 

[knowLittle]

Thank you, It's much more clear now. So, for the original problem. It would be:
(d/dx) integral from [a to x] f(t)dt =f(x)

what would "a" be in the original problem?

 

[Fredrik]

You specified that $f(x,y)=\int_y^x\cos^2 t\, dt$. Not sure I can tell you much more than that because of the forum rule that requires me to only give hints, not solutions. I suggest that you try to compute $\frac{\partial}{\partial x}f(x,y)$, using that definition of f and the formula I posted. (The f that appears in that formula is of course an arbitrary integrable function).

 

[knowLittle]

If I let "a" be "y" , then from your formula
http://texify.com/$%255Cdfrac%2520%257Bd%257D%2520%257Bdx%257D%255Cint%2520_%257By%257D%255E%257Bx%257Df%255Cleft(%2520t%255Cright)%2520dt%253Df%255Cleft(%2520x%255Cright)%2520$[/PLAIN]

would be

http://texify.com/$%255Cdfrac%2520%257Bd%257D%2520%257Bdx%257D%255Cint%2520_%257By%257D%255E%257Bx%257D%255Ccos%2520%255E%257B2%257D%255Cleft(%2520t%255Cright)%2520dt%253D%255Ccos%2520%255E%257B2%257Dx$

 

[Fredrik]

Right. Now you have almost solved the problem. Can you figure out how to compute the other partial derivative? (I have to do something else for the next hour or two, so I won't reply as fast).

You might want to check out the LaTeX guide for the forum. You can also click the quote button next to one of my posts to see how I'm doing the math. (The \, is a small space. It's probably obvious what the other codes I used represent).

 

[knowLittle]

Ok, thank you so far.

$\dfrac {d} {dy}\int _{y}^{x}\cos ^{2}\left( t\right) dt=\cos ^{2}\left( y\right)$

 

[knowLittle]

So to solve the original problem:
$f_{x}\left( x,y\right) =\cos ^{2}\left( x\right)$
and $f_{y}\left( x,y\right) =\cos ^{2}y$ ?

 

[Fredrik]

$f_{x}\left( x,y\right) =\cos ^{2}\left( x\right)$

This one is correct.

and $f_{y}\left( x,y\right) =\cos ^{2}y$ ?

This one is not. Can you figure out why?

 

[DrewD]

Edit: deleted

 

[knowLittle]

By the way, I made a mistake with the original equation in the post of today at 3:15pm.
The right equation is:
$f\left( x,y\right) =\int _{x}^{y}\cos \left( t^{2}\right) dt$
So,
$f_{x}\left( x,y\right) =\cos \left( x^{2}\right)$ is this still correct?

 

[knowLittle]

I think that this is correct:

$\dfrac {d} {dx}\int _{x}^{y}\cos t^{2}dt=-\cos x^{2}$

and

$\dfrac {d} {dy}\int _{x}^{y}\cos \left( t^{2}\right) dt=\cos \left( y^{2}\right)$

 

[Fredrik]

By the way, I made a mistake with the original equation in the post of today at 3:15pm.
The right equation is:
$f\left( x,y\right) =\int _{x}^{y}\cos \left( t^{2}\right) dt$
So,
$f_{x}\left( x,y\right) =\cos \left( x^{2}\right)$ is this still correct?

Ah, OK, I got the definition of f wrong in post #17. We have
$$\frac{\partial}{\partial y}\int_x^y\cos t^2\, dt =\cos y^2$$ and
$$\frac{\partial}{\partial x}\int_x^y\cos t^2\, dt =-\cos x^2.$$ So if $f(x,y)=\int_x^y\cos^2\, dt$, then the last equality in the quote above is wrong (but post #25 is right).

By the way, the times displayed depend on what time zone you're in.

 

[knowLittle]

Thank you. It's been great to explore mathematics and Physics Forums with you. :)