Question

evaluate d/dx ∫x,5] (t^5)tan(t)dt

Answer 1

per the usual rules, with ∫[lower .. upper],

d/dx ∫[5 .. x] (t^5) tan(t) dt = x^5 tanx

Answer 2

this is just the Chain Rule in reverse. For, if

F(t) = ∫f(t) dt then if u and v are functions of x,
∫[u..v] f(t) dt = F(v) - F(u)
so
d/dx ∫[u..v] f(t) dt = F'(v) dv/dx - F'(u) du/dx = f(v) dv/dx - f(u) du/dx