Use the fundamental theorem of calculus to find the derivative of the function?

#h(x) = int_1^(e^x)3ln(t)dt#

h'(x) = ?

1 Answer
Steve M
Dec 24, 2017

# d/dx \ int_1^(3x) \ 3lnt \ dt = 3xe^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:

# h'(x) # where #h(x) = int_1^(e^x) \ 3lnt \ dt#

ie, we want:

# h'(x) = d/dx int_1^(e^x) \ 3lnt \ 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=e^x => (du)/dx = e^x #

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

# d/dx \ int_1^(e^x) \ 3lnt \ dt = d/dx \ int_1^u \ 3lnt \ dt #

# " " = (du)/dx*d/(du) \ int_1^u \ 3lnt \ dt #

# " " = e^x \ d/(du) \ int_1^u \ 3lnt \ dt #

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

# d/dx \ int_1^(e^x) \ 3lnt \ dt = e^x(3lnu) #

And restoring the initial substitution we get:

# d/dx \ int_1^(3x) \ 3lnt \ dt = 3e^xlne^x = 3xe^x#

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