Is it even do-able to take the following derivative $$\frac{d\Big(\int_{0}^t f(i,t)\ di\Big)}{dt}$$
if yes, how?
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what does the $\sum_i^t$ notation stand for? – Vim Sep 09 '16 at 01:04
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summation from $i=0$ to $i=t$ – user3639557 Sep 09 '16 at 01:05
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1then $t\in\Bbb N$ and how do you take derivative w.r.t. a discrete variable? – Vim Sep 09 '16 at 01:07
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you are right. I was imprecise in defining the notation. – user3639557 Sep 09 '16 at 01:07
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@user3639557 well, if defined correctly, you can differentiate with respect to a "sum". Set it equal to a differentiable function equal to the sum for all $t\in\mathbb N$. – Simply Beautiful Art Sep 09 '16 at 01:10
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In what you've written, $t$ is a dummy variable of integration; it's meaningless to differentiate with respect to it. Are you asking how to compute something like $\frac{\partial}{\partial t} \int f(t, s), ds$? – anomaly Sep 09 '16 at 01:10
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1It's still meaningles. Do you mean $\frac{d}{dt}\int_{0}^{t} f(s,t),ds$? There is no reason for partial derivatives in this case. – Thomas Andrews Sep 09 '16 at 01:11
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When you differentiate with respect to the integral, you problem want to use the FTOC for it to make sense. – Simply Beautiful Art Sep 09 '16 at 01:11
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@ThomasAndrews yes - was just fixing the notation. So this derivative is equal to $\int^{t}_{0}f(i,t)di$? Note that $t$ explicitly appears in $f(i,t)$ – user3639557 Sep 09 '16 at 01:17
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This 2017 post now answers your question: How to compute $\frac{d}{dt} \int_{a(t)}^{b(t)} f(x,t) dx$. It was itself a duplicate of this one and that one, both of 2011 (and probably many others). – Anne Bauval Mar 29 '23 at 12:27
1 Answers
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Let $g(t)=\int_{0}^{t} f(s,t)\,ds$.
Then $$g(t+h)-g(t)= \int_{0}^{t} (f(s,t+h)-f(s,t))\,ds + \int_{t}^{t+h} f(s,t+h)\,ds$$
So $$\frac{g(t+h)-g(t)}{h} = \int_{0}^{t} \frac{f(s,t+h)-f(s,t)}{h}\,ds + \frac{1}{h}\int_{t}^{t+h} f(s,t+h)\,ds$$
If $f$ is "nice", then the limit of this as $h\to 0$ is:
$$\frac{dg}{dt} = \int_0^t \frac{\partial f}{\partial t}(s,t)\,ds + f(t,t)$$
I forget the conditions under which you can switch the limit and integral, but it is true for lots of $f$.
The second term $f(t,t)$ really requires $f$ to be continuous, at least on $t.$
Thomas Andrews
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In general we have $dg(t,y)/dt=(\partial g/\partial t)+(\partial g/\partial y)(dy/dt).$ In this case let $g(t,y)=\int_0^yf(t,i)di.$ – DanielWainfleet Sep 09 '16 at 02:00