Derivative is taken at a point and hence is value at a point. But definite integral is the value over a domain. Then how come derivative of definite integral make sense.
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The interval and the value over the interval can depends from a variable, so we can define derivative with respect to this variable. – Emilio Novati Dec 29 '17 at 13:57

It makes as much sense as taking the square root and then squaring the answer. – For the love of maths Dec 29 '17 at 13:58

3Please add an example. I think you are referring to something like $\frac{d}{dx} \int_{1}^{2} t^x dt$. Here there is another variable $x$. – jonsno Dec 29 '17 at 14:02

Yes. $x$ and $t$ are independent variables – lorilori Dec 29 '17 at 15:10
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Take a function $f$. Fix a point in its domain, call it $a$. We would like to know what the integral of this function is for different values, i.e. we have a function of $x$ defined by:
$$F(x) = \int_a^xf(t)dt$$
Note that this is a function of the upper limit of integration.
The fundamental theorem of calculus states that the derivative of this function at a given point $x$ is the value of $f$ evaluated at $x$, i.e.
$$F'(x) = \frac{d}{dx} \int_a^xf(t)dt = f(x) $$
JuliusL33t
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If one or both limits are variable then the definite integral is a function and in some cases the derivative does exist. If both limits are constant then the definite integral is a constant therefore the derivative is zero.
Mohammad RiaziKermani
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