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How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function ##h(x) = int arctan t dt## from [2,1/x]?

The theorem says:

Given the function:

##y=int_(h(x))^g(x)f(t)dt##

then:

##y’=f(g(x))*g'(x)-f(h(x))*h'(x)##.

So if:

##y=int_0^(1/x)arctantdt##,

then

##y’=arctan(1/x)*(-1/x^2)-arctan2*0=##

##=-arctan(1/x)/x^2##.

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