What is the derivative of #int_(cosx)^(7x) \ cost^3 \ dt # wrt #x#?

1 Answer
Jul 25, 2017

Answer:

# d/dx int_(cosx)^(7x) \ cost^3 \ dt = 7 cos(7x)^3 + sinx cos(cos^3x) #

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:

# g'(x) = d/dx \ int_(cosx)^(7x) \ cost^3 \ dt#

Note that neither the upper or lower bounds are in the correct format for the FTOC to be applied, directly. We can manipulate the definite integral as follows:

# int_(cosx)^(7x) \ cost^3 \ dt = int_(a)^(7x) \ cost^3 \ dt - int_a^(cosx) \ cost^3 \ dt #

And so:

# g'(x) = d/dx int_(a)^(7x) \ cost^3 \ dt - d/dx int_a^(cosx) \ cost^3 \ dt # ... [A]

We have arbitrary chosen the lower limit as some arbitrary constant #a# (we could use #0# wlog, any number will do!). We can further manipulate these definite integral using a substitution and the chain rule. Let:

# u=7x => (du)/dx = 7 #
# v=cosx => (dv)/dx = -sinx #

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

# g'(x) = d/dx int_(a)^u \ cost^3 \ dt - d/dx int_a^v \ cost^3 \ dt #

# " " = (du)/dx d/(du) int_(a)^u \ cost^3 \ dt - (dv)/dx d/(dv) int_a^v \ cost^3 \ dt #

# " " = 7 d/(du)int_(a)^u \ cost^3 \ dt - (-sinx) d/(dv) int_a^v \ cost^3 \ dt #

# " " = 7 d/(du)int_(a)^u \ cost^3 \ dt + sinx d/(dv) int_a^v \ cost^3 \ dt #

And now the derivative of both the integrals are in the correct form for the FTOC to be applied, giving:

# g'(x) = 7 cosu^3 + sinx cosv^3 #

And restoring the initial substitution we get:

# g'(x) = 7 cos(7x)^3 + sinx cos(cos^3x) #