Question

(2x^2+4)(3x+3) over 7x-6 Find the derivative using quotient rule Help please!

Answer 1

If I were you I would multiply that out first, but let me see something first, hold on....

Answer 2

yeah multiply out in the numerator first will make your life much easier. you do not want to use the quotient rule combined with the product rule if you can avoid it

Answer 3

i will be quiet now

Answer 4

f'(x)=\[=((2x^2+4)(3x+3))'(7x-6)-(2x^2+4)(3x+3)(7x-6)' \over (7x-6)\]\[=((2x^2+4)'(3x+3)+(2x^2+4)(3x+3)')(7x-6)-(2x^2+4)(3x+3)(7x-6)' \over (7x-6)\]

Answer 5

Sorry. Over (7x-6)^2

Answer 6

Wow, you're really taking the long way on this one

Answer 7

Thank you. Saved me hella typing.

Answer 8

hmmm

Answer 9

You smell. There. I said it. Now satellite doesn't have to.

Answer 10

\[(2x^2+4)(3x+3)=6 x^3+6 x^2+12 x+12\]

Answer 11

I hate this shiz

Answer 12

sorry, just thought i would mention it

Answer 13

i got that im not sure about my final anwer

Answer 14

your first step should look like
\[\frac{(7x-6)(18x^2+12x+12)-(6x^3+6x^2+12x+12)7}{(7x-6)^2}\] then a raft of annoying ugly algebra

Answer 15

can yiu show me the rest..

Answer 16

sprusty, don't let me catch you making a mistake, you will get the wrath from me.

Answer 17

this one? do i really have to do this algebra in the numerator? it is multiply and combine like terms

Answer 18

\[ \frac{6 (14 x^3-11 x^2-12 x-26)}{(7x-6)^2}\]there factored and all

Answer 19

thank you so much

Answer 20

yw

Answer 21

can you help me on the other ones?

Answer 22

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e83ec8a0b8bc11dd5526e18