What Does Differentiating Both Sides as Integrals of the Upper Limits Mean?

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SUMMARY

The discussion centers on the differentiation of integrals with respect to their upper limits, specifically in the context of the fundamental theorem of calculus and the Leibniz rule. The equation X'(t)/a(X(t)) = b(t) illustrates the application of these concepts, where the left-hand side represents the derivative of an integral with variable upper limits. The example provided demonstrates that the derivative of the integral f(x) = ∫_{a}^{x} u(t) dt results in f'(x) = u(x), confirming the relationship between differentiation and integration as established by the fundamental theorem of calculus.

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TonyAlmeidaAtLSE
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Hi, I know I sound the most stupid here but I have an easy question here please help out:

X'(t)=dX(t)/dt

If dX/a(x) = b(t)dt
∫d(x)/a(x) = ∫b(s) ds {The upper and lower limites of lefthand side are X(t) and X(0) respectively and those of righthand side are t and 0}

Then differentiating both sides as integrals of the upper limits
X’(t)/a(X(t)) =b(t)
I don’t understand what does “differentiating both sides as integrals of the upper limits” mean? Here does it have anything to do with Leibniz Rule? What about lower limits X(0)

Your generous help and reply will be greatly appreciated.
 
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Consider the example:
[tex]f(x)=:\int_{a}^{x} u(t) dt[/tex] (1)
Take the primitive (antiderivative) of u(t) to be U(t).Then,equation (1) becomes,after applying the fundamental formula of Leibniz & Newton
[tex]f(x)=U(x)-U(a)[/tex] (2)
These 2 functions differ through a constant,namely U(a).So it's natural to conclude they have the same derivative.
[tex]f'(x)=U'(x)=u(x)[/tex] (3)
,where i made use of the fact that U is the antiderivative of u,no matter whether the latter's argument is "x","t","v","y",...
Formula (3) means
[tex]\frac{d}{dx}\int_{a}^{x} u(t) dt=u(x)[/tex]
,where "a" is a constant.

Daniel.
 
Although it is a special case of Leibniz rule, it is really just the "fundamental theorem of calculus".
 

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