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