How do you integrate #f(x)=(2-2x+2x^2-2x^3-3x^4)(1+x-2x^2-3x^3)# using the product rule?

1 Answer
Nov 8, 2016

Answer:

#(2-2x+2x^2-2x^3-3x^4) (1 - 4x - 9x^2) + (-2 + 4x - 6x^2 - 12x^3) (1 + x -2x^2 -3x^3)#

Explanation:

Remember the product rule is f g'+f' g, so:
#(2-2x+2x^2-2x^3-3x^4) (1 - 4x - 9x^2) + (-2 + 4x - 6x^2 - 12x^3) (1 + x -2x^2 -3x^3)#

You are taking the derivative of f and that what is g. When you take the derivative of #(2-2x+2x^2-2x^3-3x^4)# you are multiplying by the exponent. Step by step for #(2-2x+2x^2-2x^3-3x^4)# is #2^0#, which is 0. the 2x is #2x^1#, where you multiply by the 1, where you get the 2, and take away the x. If you go down the line, for each and everyone one it will fall the exact same. #3x^4# become #12x^3#. Remember you are multiplying by the exponent and then subtracting it by 1.

If someone were to give you find the derivative of #(2x^5-2x^2-5) (3x^2+2)#, what would you write?

Starting from the beginning one step at a time.

So 1, Formula: f g' + f' g

2, find the derivatives:
#(2x^5-2x^2-5)# becomes #(10x^4-4x)#
#(3x^2+2)# becomes (6x)

3, Now put in what you have:
#(2x^5-2x^2-5)# #(6x)# + #(10x^4-4x)##(3x^2+2)#

Super easy right?