Michael J. answered • 10/01/15
Tutor
5
(5)
Mastery of Limits, Derivatives, and Integration Techniques
f(x) = (2x - 3)4(x2 + x + 1)5
We use a combination of product rule and chain rule.
f'(x) = [2 * 4(2x - 3)3(x2 + x + 1)5] + [(2x - 3)4 * 5(x2 + x + 1)4(2x + 1)]
f'(x) = 8(2x - 3)3(x2 + x + 1)5 + (2x - 3)45(x2 + x + 1)4(2x + 1)
We can take out a common factor. That common factor is (2x - 3)3(x2 + x + 1)4.
f'(x) = (2x - 3)3(x2 + x + 1)4[8(x2 + x + 1) + 5(2x - 3)(2x + 1)]
As for the rest, I leave up to you to simplify even further.
We use a combination of product rule and chain rule.
f'(x) = [2 * 4(2x - 3)3(x2 + x + 1)5] + [(2x - 3)4 * 5(x2 + x + 1)4(2x + 1)]
f'(x) = 8(2x - 3)3(x2 + x + 1)5 + (2x - 3)45(x2 + x + 1)4(2x + 1)
We can take out a common factor. That common factor is (2x - 3)3(x2 + x + 1)4.
f'(x) = (2x - 3)3(x2 + x + 1)4[8(x2 + x + 1) + 5(2x - 3)(2x + 1)]
As for the rest, I leave up to you to simplify even further.
