How Do You Differentiate an Integral with a Time-Dependent Upper Limit?

[moo5003]

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 don't 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.

 

[Hurkyl]

Er,

$$\frac{d}{dx} e^{u(x)} \neq e^{\frac{d}{dx} u(x)$$

(usually)

 

[Hurkyl]

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.

 

[moo5003]

Er,

$$\frac{d}{dx} e^{u(x)} \neq e^{\frac{d}{dx} u(x)$$

(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

 

[moo5003]

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