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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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##.