- #1

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[tex] (f*g)^\prime = \frac{d}{dx} \int f(x) g(x-u) dx = \int f(x) \frac{d}{dx} g(x-u) dx = f(x) * g(x)^\prime [/tex]

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- Thread starter radiator
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- #1

- 23

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[tex] (f*g)^\prime = \frac{d}{dx} \int f(x) g(x-u) dx = \int f(x) \frac{d}{dx} g(x-u) dx = f(x) * g(x)^\prime [/tex]

- #2

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$$(f*g)^\prime(x) = \frac{d}{dx} \int f(u) g(x-u) du = \int f(u) \frac{d}{dx} g(x-u) du = (f * g^\prime)(x)$$

This is a special case of a more general situation. Suppose

$$F(x) = \int_{a}^{b} f(x,y) dy$$

where ##a## and ##b## are constants not dependent on ##x##. Then if ##f## is sufficiently nice, we have

$$F'(x) = \int_{a}^{b} \frac{\partial}{\partial x} f(x,y) dy$$

Sufficiently nice means, for example, that ##f## is continuous and that ##\partial f /\partial x## exists and is continuous. No point writing out a proof here; see this Wikipedia entry:

http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign#Proof_of_Theorem

- #3

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Thanks for the correction. I guess I understand the justification.

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