Chain rule question

by bitrex
Tags: chain, rule
bitrex is offline
Nov26-09, 08:21 PM
P: 196
I'm having some trouble following this equation:

[tex]\frac {d \Phi_B} {dt} = (-) \frac {d}{dx_C} \left[ \int_0^{\ell}dy \ \int_{x_C-w/2}^{x_C+w/2} dx B(x)\right] \frac {dx_C}{dt} = (-) v\ell [ B(x_C+w/2) - B(x_C-w/2)] \ [/tex]

Shouldn't the differentiation of the bracketed terms "killed" the integration of B(x)? Why is it still evaluated between the limits in the final expression? Thanks for any advice.
Phys.Org News Partner Science news on
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
CompuChip is offline
Nov27-09, 09:16 AM
Sci Advisor
HW Helper
P: 4,301
There is a theorem in calculus that
[tex]\frac{d}{dx} \int_0^x f(y) \, dy = f(x) [/tex]
(i.e. differentiating with respect to the integral boundary)
which you should be able to find in your calculus book.

You can prove your identity by taking any fixed point [itex]x_C - w/2 < x_0 < x_C + w/2[/itex] and write
[tex] \int_{x_C - w/2}^{x_C + w/2} dx B(x) = \int_{x_C - w/2}^{x_0} dx B(x) + \int_{x_0}^{x_C + w/2} dx B(x) = \int_{x_0}^{x_C + w/2} dx B(x) - \int_{x_0}^{x_C - w/2} dx B(x)[/tex]
bitrex is offline
Nov28-09, 01:21 AM
P: 196
Thank you! That's a good theorem to know!

HallsofIvy is offline
Nov28-09, 06:20 AM
Sci Advisor
PF Gold
P: 38,879

Chain rule question

Yes, it's called the Fundamental Theorem of Calculus!
bitrex is offline
Nov28-09, 03:57 PM
P: 196
Well, I suppose if I can't recognize that, this is all kind of a lost cause!
elibj123 is offline
Nov29-09, 01:30 PM
P: 240
actually a more general theorem is:


Register to reply

Related Discussions
chain rule question Calculus & Beyond Homework 0
Derivatives Chain Rule question Calculus & Beyond Homework 9
chain rule question Calculus & Beyond Homework 2
Chain Rule/Product Rule Question Calculus & Beyond Homework 2
Chain Rule Question Introductory Physics Homework 4