Question #16171
1 Answer
May 12, 2017
# f'(x) = 2(2x-1)^2(x^2+3) (7x^2-2x+9) #
Explanation:
We can use a combination of the chain rule and the product rule:
We have:
# f(x) = (2x-1)^3(x^2+3)^2 #
We can use the chain rule on each individual function. Let
# { (u=(2x-1)^3), (v=(x^2+3)^2) :} => { ((du)/dx=3(2x-1)^2(2),=6(2x-1)^2), ((dv)/dx=2(x^2+3)(2x),=4x(x^2+3)) :} #
And then:
# f(x) = uv #
And by the product rule we have:
# f'(x) = (u)((dv)/dx) + ((du)/dx)(v) #
# " " = (2x-1)^3 4x(x^2+3) + 6(2x-1)^2(x^2+3)^2) #
# " " = (2x-1)^2(x^2+3) {4x(2x-1)+6(x^2+3)} #
# " " = (2x-1)^2(x^2+3) (8x^2-4x+6x^2+18) #
# " " = (2x-1)^2(x^2+3) (14x^2-4x+18) #
# " " = 2(2x-1)^2(x^2+3) (7x^2-2x+9) #
