$$\frac d{dt}\int_0^t f(t,\tau)d\tau=f(t,t)+\int_0^t\frac\partial{\partial t}f(t,\tau)d\tau$$
How would you prove that? Thank you very much
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You would replace $\tau$ with $t$ and then bring the derivative inside the integral (so that it becomes a partial derivative) per the Newton-Leibniz integral rule. – Axion004 Nov 27 '19 at 17:41
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1Possible duplicate of when can we interchange integration and differentiation – Eric Towers Nov 27 '19 at 21:02
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By the Newton-Leibniz Integral rule
$$\frac{d}{dt}\int_a^{s(t)}f(t,\tau)d\tau=s'(t)\, f(s(t),t)+\int_a^{s(t)}\frac{\partial f(t,\tau)} {\partial t } \, d\tau$$
in your case $s(t)=t$ and $a=0$. Therefore, $s'(t)=1$ and the above expression becomes
$$\frac{d}{dt}\int_0^{t}f(t,\tau)d\tau=\, f(t,t)+\int_0^{t}\frac{\partial f(t,\tau)} {\partial t } \, d\tau$$
Axion004
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Thank you very much, do you think is possible to show that by the definition of derivative ? I am afraid we are not allowed to use NL rule. – mmm Nov 27 '19 at 18:29
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Use the Leibniz integral rule.
Or use any of the other answers to this and variants of this question on this site.
Eric Towers
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