How do you use the second fundamental theorem of Calculus to find the derivative of given #int (cos(t^3))# from #[cosx, 7x]#?

1 Answer
Jul 9, 2016

Answer:

# 7 cos (7x)^3 + sin x cos (cos^3 x) #

Explanation:

FTC part 2 tells us that

#color{blue}{d/(du) int_a^u \ f(t) \ dt = f(u)}# where #a = const#

if #u = u(x)#, we can add in the chain rule to conclude that

#d/dx int_a^u \ f(t) \ dt = f(u)* (du)/dx#

we can also further conclude from FTC that

#d/(du) int_u^a \ f(t) \ dt = - d/(du) int_a^u \ f(t) \ dt = - f(u)#

if we put that all together we get from the FTC and the chain rule:

#d/(dx) int_u^v \ f(t) \ dt = f(v)* (dv)/dx - f(u) * (du)/dx#

so here we have

#f(t) = cos t^3#

# u = cos x, u' = -sin x#
#v = 7x, v'= 7#

and so we pattern match to get

# 7 cos (7x)^3 + sin x cos (cos^3 x) #