Integrating [itex]\frac{\partial f(x,y)}{\partial y(x)}[/itex] with respect to x

[nasshi]

I don't know how to apply Leibniz's rule for this integrand. I believe there is a substitution to allow me to express this integral differently, but the partial being a function of what we're integrating by is confusing me.

\int^{b}_{a} \frac{\partial f(x,y)}{\partial y(x)} dx

Knowns are that [a,b] is a subset of the domain of x and that y is a function of x.

Can someone please suggest something to get me started, or ask me a question that may get my brain going in the right direction?

I attempted working with \int^{b}_{a} \frac{\partial f(x,y)}{\partial x} \frac{dx}{dy}dx by expanding the partial according to the chain rule, but I didn't think it helped me.

Edit: I didn't think the above would help because what exactly does [f(b,y) - f(a,y)]\frac{dx}{dy} mean? I don't know how to interpret the factor \frac{dx}{dy}.

 

[HallsofIvy]

I don't understand your notation. \partial f/\partial y is defined as the derivative of y, treating x as a constant. With that definition, \partial f/\partial y(x) is meaningless. If y is a function of x and x is a constant, then y is a constant, not a variable and we cannot differentiate with respect to it. And the problem you are working on, \int_a^b\frac{\partial f}{\partial x}\frac{dx}{dy}dx does not imply that. You are, I think, using the chain rule incorrectly. You want, of course, "df" so that you can integrate. But the chain rule gives df= \left(\frac{\partial f}{\partial x}\frac{dx}{dy}+ \frac{\partial f}{\partial y}\right)dy, not at all what you have- in addition to an additional term, you end with "dx", not "dy".

The fundamental problem is that your integral, \int_a^b\frac{\partial f}{\partial x}\frac{dx}{dy}dx has an "extra" \frac{1}{dy} that just doesn't make sense in an integral.

 

[jackmell]

I don't know how to apply Leibniz's rule for this integrand.

You can apply it but I don't think it will make much difference. Note:

\frac{\partial}{\partial y}\int_a^b f(x,y)dx=\int_a^b \frac{\partial f}{\partial y} dx

 

[nasshi]

My original equation was taken from something with messy notation, and I lost some subscripts while working part of the larger question out. So I gave you a poorly framed question to answer -- I'm sorry.

In my original question, the function f(x,y) is vector valued, and the integrand is with respect to dx_{1}[.

Since I just noticed how sloppy my notation is getting, I'll go back through my problem with more precision before I ask another question.

 

[jackmell]

My original equation was taken from something with messy notation, and I lost some subscripts while working part of the larger question out. So I gave you a poorly framed question to answer

. . . then I want my money back.

 

[nasshi]

. . . then I want my money back.

Don't worry, it's... uh... already in the mail!