# Chain rule question

by bitrex
Tags: chain, rule
 P: 196 I'm having some trouble following this equation: $$\frac {d \Phi_B} {dt} = (-) \frac {d}{dx_C} \left[ \int_0^{\ell}dy \ \int_{x_C-w/2}^{x_C+w/2} dx B(x)\right] \frac {dx_C}{dt} = (-) v\ell [ B(x_C+w/2) - B(x_C-w/2)] \$$ Shouldn't the differentiation of the bracketed terms "killed" the integration of B(x)? Why is it still evaluated between the limits in the final expression? Thanks for any advice.
 Sci Advisor HW Helper P: 4,301 There is a theorem in calculus that $$\frac{d}{dx} \int_0^x f(y) \, dy = f(x)$$ (i.e. differentiating with respect to the integral boundary) which you should be able to find in your calculus book. You can prove your identity by taking any fixed point $x_C - w/2 < x_0 < x_C + w/2$ and write $$\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)$$
 P: 196 Thank you! That's a good theorem to know!
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PF Gold
P: 38,879

## Chain rule question

Yes, it's called the Fundamental Theorem of Calculus!
 P: 196 Well, I suppose if I can't recognize that, this is all kind of a lost cause!
 P: 240 actually a more general theorem is: $$\frac{d}{dx}\int^{b(x)}_{a(x)}f(s)ds=f(b(x))\frac{db}{dx}-f(a(x))\frac{da}{dx}$$

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