Question

How do you calculate the derivative of `int(6t + sqrt(t)) dt` from `[1, tanx]`?

Answer 1

Use the Fundamental theorem of Calculus Part ! and the chain rule.

Explanation:

`y = int_1^u f(t) dt` then `dy/(du) = f(u)`

In this case we want `dy/dx` so we need the chain rule: `dy/dx = dy/(du) (du)/dx`

`y = int_1^tanx (6t+sqrtt) dt`

`dy/dx = underbrace([6(tanx)+sqrt(tanx)])_((dy)/(du))* underbrace(sec^2x)_((du)/dx)`

`=(6tanx+sqrt(tanx))sec^2x`