Answer:
#15: 1/5 (x^2 - x - 3)^5
#18: 5 (3x^2 - 6x + 1)^5
Step-by-step explanation:
#15
(2x-1)(x^2-x+3)^4
Chain Rule!!!
g'(x)f'(g(x)) = d/dx(f(g(x)) = (f ⚬ g)'(x)
f'(x) = x^4
g(x) = x^2 - x - 3
g'(x) = 2x - 1
So we only have to integrate x^4 gives 1/5 x^5
f(x) = 1/5 x^5
f(g(x)) = (f ⚬ g)(x) = 1/5 (x^2 - x - 3)^5
#18
150(x-1)(3x^2-6x+1)^4
Chain rule again!! In disguise.
f'(x) = x^4
f(x) = 1/5 x^5
g(x) = 3x^2-6x+1
g'(x) = 6x - 6 = 6(x-1)
150(x-1) = 25 g'(x)
25 g'(x)f'(g(x)) = 25 (f ⚬ g)(x) = 5(3x^2-6x+1)^5
