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Math Help - derivative with x occurring at both integrand and upper bound

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    derivative with x occurring at both integrand and upper bound

    If f is continuously differentiable in \mathbb R^2, g is continuously differentiable in \mathbb R, a a constant real, how to calculate the derivative of \int_a^{g(x)}f(x,t)dt with regard to x (not by numerical methods)? Thanks.
    Last edited by zzzhhh; November 25th 2009 at 08:15 AM.
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    Quote Originally Posted by zzzhhh View Post
    If f is continuously differentiable in \mathbb R^2, g is continuously differentiable in \mathbb R, a a constant real, how to calculate the derivative of \int_a^{g(x)}f(x,t)dt with regard to x (not by numerical methods)? Thanks.

    Use Leibnitz's Rule for differentiation under the integral sign:

    \frac{\partial}{\partial x}\int\limits_a^{g(x)}f(x,t)\,dt=\int\limits_a^{g(  x)}\frac{\partial f(x,t)}{\partial x}\,dx+f(x,g(x))\frac{\partial g(x)}{\partial x}-f(x,a)\frac{\partial a}{\partial x}

    Tonio
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    Thank you, erudite Tonio.
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    After reading the proof of this general Leibniz integral rule, I have to stress that the conditions supporting the special case of Leibniz integral rule(*) must be preserved to make possible the interchange of the lim and int process in producing the integral part of the result when \Delta x\to 0, the most important of which is the continuity of the partial derivative.
    (*) \frac{d}{dx}\int_a^b f(x,t)dt=\int_a^b \frac{\partial}{\partial x}f(x,t)dt where a,b are both constant.
    ps: My question comes from the proof of Poincare's lemma in baby Rudin, the proof that F\in\mathcal C'(E) is omitted as usual in Th10.38, I think the general Leibniz integral rule is necessary to prove it which is not mentioned at all in this book. I hope it is not the author's original intention to leave such a huge gap for me to fill. Thanks to Tonio I finally made it.
    Last edited by zzzhhh; November 26th 2009 at 01:56 PM.
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