Differntiation on Integrals

  • Thread starter moo5003
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  • #1
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Main Question or Discussion Point

How would you differntiate the following type of formula:


d/dt [ e^(Integral from 0 to t [v(x+s,t-s) ds])f(x+t)]


Note: I dont want to change variables, I know I can make it nicer but I would like to know how to do it directly.



So far:

= e^(d/dt[integral from 0 to t(v(x+s,t-s)ds)]) * f(x+t) + e^(integral from 0 to t[v(x+s,t-s)ds]) * d/dt(f(x+t))

So I really just need help doing:

d/dt [ Integral from 0 to t (v(x+s,t-s)ds)]

= v(x+s,t-s) right? I'm not sure about this.
 

Answers and Replies

  • #2
Hurkyl
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Er,

[tex]\frac{d}{dx} e^{u(x)} \neq e^{\frac{d}{dx} u(x)[/tex]

(usually)
 
  • #3
Hurkyl
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So I really just need help doing:

d/dt [ Integral from 0 to t (v(x+s,t-s)ds)]

= v(x+s,t-s) right? I'm not sure about this.
Nope. Apply Leibniz's rule. Or derive it yourself by using limits.
 
  • #4
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Er,

[tex]\frac{d}{dx} e^{u(x)} \neq e^{\frac{d}{dx} u(x)[/tex]

(usually)
True, though I still need to find out d/dx[u(x)] or in my case what I listed above. Since d/dt(e^u(x,t)) = d/dt[u(x,t)]e^u(x,t)


Leibniz's rule:

Since I'm unsure v or d/dt are continuous on over the region does that mean I cannot pass the differentiation?

Or can I still use:

d/dt[Integral from a to b(f(x,t)dx)] =

Integral from a to b[d/dt(f(x,t)dx)] + f(b,t)db/dt - f(a,t)da/dt
 
Last edited:
  • #5
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So, I think I did it but I'm unsure and I would like you guys to confirm it if possible.

d/dt[ int 0 to t ( V(x+s,t-s)ds )
=
Int 0 to t [ d/dt ( v(x+s,t-s) ) ] + v(x+t, 0)dt/dt - v(x,t)d0/dt
=
Into 0 to t [ d/dt ( v(x+s,t-s) ) + v(x+t,0)

Assuming continuity over the region
 

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