Question

How do you find `F'(x)` given `F(x)=int 1/t dt` from `[1,3x]`?

Answer 1

`d/dx \ int_1^(3x) \ 1/t \ dt = 1/x`

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

`d/dx \ int_a^x \ f(t) \ dt = f(x)` for any constant `a`

(ie the derivative of an integral gives us the original function back).

We are asked to find:

`F'(x)` where `F(x) = int_1^(3x) \ 1/t \ dt`

ie, we want:

`F'(x) = d/dx int_1^(3x) \ 1/t \ dt` ..... [A}

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

`u=3x => (du)/dx = 3`

The substituting into the integral [A], and applying the chain rule, we get:

`d/dx \ int_1^(3x) \ 1/t \ dt = d/dx \ int_1^u \ 1/t \ dt`

`" " = (du)/dx*d/(du) \ int_1^u \ 1/t \ dt`

`" " = 3 \ d/(du) \ int_1^u \ 1/t \ dt`

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

`d/dx \ int_1^(3x) \ 1/t \ dt = 3 \ 1/u`

And restoring the initial substitution we get:

`d/dx \ int_1^(3x) \ 1/t \ dt = 3 \ 1/(3x) = 1/x`