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

1 Answer
Jun 20, 2017

Answer:

# 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#