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Fundamential theorem of calculus says $$\frac{d}{dx} \int_a^x f(t) dt = f(x).$$

I was wondering if $$\frac{d}{dx} \int_a^x f(x,t) dt = f(x,x)?$$

Is this mentioned in Baby Rudin book or somewhere else?

Thanks and regards!

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If you consider the simple case $f(x,t)= x$, you will see that the relation is not true. – Fabian Apr 11 '11 at 21:33
@Fabian: How is it not true? I think the contrary? – Tim Apr 11 '11 at 21:44
calculate $\int_a^x f(x,t) dt = \int_a^x x dt = x (x-a)$. Then you see that $\partial_x \int_a^x f(x,t) dt = \partial_x x (x-a) = 2x -a \neq f(x,x) = x.$ – Fabian Apr 11 '11 at 21:47
up vote 6 down vote accepted

See the wikipedia page on Differentiation under the integral sign - we have that $$\frac{d}{dx} \int_a^x f(x,t) dt = f(x,x)\cdot1-f(x,a)\cdot 0 + \int_a^x\frac{\partial}{\partial x} f(x,t)dt=f(x,x) + \int_a^x\frac{\partial}{\partial x} f(x,t)dt$$

so that $\frac{d}{dx} \int_a^x f(x,t) dt = f(x,x)$ only holds for all $x$ if and only if $\frac{\partial}{\partial x} f(x,t)=0$ a.e. (I think?), which (metaphorically) means $f$ was really only a function of $t$ all along.

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