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]?

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Answer 1 · 2015-05-14T11:56:29

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#.