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Mathematics 18 Online
OpenStudy (anonymous):

Calculate dy/dx You need not expand your answer. y=3x^2-9x+13/2x+4

OpenStudy (anonymous):

dy/dx=

OpenStudy (anonymous):

i got 3x^2+8x=26 which is clearly wrong

OpenStudy (anonymous):

over 2(x+2)^2

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

gimmie a sec

OpenStudy (anonymous):

kk thanks for helping

OpenStudy (anonymous):

gah I hate quotient rule

OpenStudy (anonymous):

the answer is 3x^2-9x+13/(2x+4)^2

OpenStudy (anonymous):

now I will put my work on here

OpenStudy (anonymous):

Do you know the quotient rule? \[(u\div v)^'=(vu^'-uv')\div v^2\]

OpenStudy (anonymous):

i got that too agentzero and it was wrong :/

OpenStudy (anonymous):

actually 3x^2+12x-31/2(2x+4)^2

OpenStudy (anonymous):

\[u = 3x{^2}-9x+13\]\[v = 2x+4\] \[u' = 6x - 9\]\[v' = 2\] \[y' = [((2x+4)\times(6x-9)) - ((3x{^2} - 9x + 13)\times 2)]\div (2x+4){^2}\]

OpenStudy (anonymous):

thats what wolfram says. This isnt hard I dont know why im getting it wrong

OpenStudy (anonymous):

think i gotta go to a tutor for this one lol. thanks for the help

OpenStudy (anonymous):

(2x+4)(6x-9)-(2)(3x^2-9x+13) (12x^2+6x-36)-(6x^2-18x+26) 6x^2+24x-62/(2x+4)^2 6x^2+24x-62/4x^2+16x+16 factor a 2 out 3x^2+12x-31/2x^2+8x+8 = 3x^2+12x-31/2(x+4)^2

OpenStudy (anonymous):

if thats not right then i dont know what is lol

OpenStudy (anonymous):

I got the same answer twice

OpenStudy (anonymous):

I just gave it to you... \[y' = [((2x+4)\times(6x-9)) - ((3x{^2} - 9x + 13)\times 2)]\div (2x+4){^2}\] \[(2x+4)(6x-9) - 2(3x^2 - 9x + 13) \over (2x+4)^2\] \[(12x^2 - 18x + 24x - 36) - (6x^2 - 18x + 26) \over (2x+4)^2\] \[(12x^2 + 6x - 36) - (6x^2 - 18x + 26) \over (2x+4)^2\] \[6x^2 + 24x - 62 \over (2x+4)^2\] You can simplify it further but it's not necessary: \[3x^2 + 12x - 31 \over 2(2x+4)^2\]

OpenStudy (anonymous):

There's three basic rules you MUST know for differentiation: The product rule: \[(uv)' = u'v + uv'\] The quotient rule: \[(u\div v)' = {(vu' - uv')\over v{^2}}\] The chain rule: \[f[g(x)]^'=f^' [g(x)]\times g(x)\]

OpenStudy (anonymous):

@AgentZero: for problems you are given (i.e. not "real world" problems) you generally do not need to mess with the denominator. Once you have (vu' - uv')/v^2 you can generally just simplify the numerator and stop there.

OpenStudy (anonymous):

sure but Im assuming this is computer homework and it would probably want denominator too

OpenStudy (anonymous):

3x^2+12x−31/2(2x+4)2 yeah its for webassign. I the answer you both got on first try but it still doesnt accept it

OpenStudy (anonymous):

i got the same answer you both got is what im saying* but webassign didnt accept it..

OpenStudy (anonymous):

I have a lot of experience with webassign and I can say it is terrible =)

OpenStudy (anonymous):

to fewscrewsmissing your answer is good except the last denominator it should be 2(x+2)^2

OpenStudy (anonymous):

he did in his last answer

OpenStudy (anonymous):

you probably didn't pay attention inside of () is x+2

OpenStudy (anonymous):

Yes you're right MarinaDL, thanks for pointing that out. I would normally stop prior to this step anyway. Apparently this particular assignment is requiring very specific input.... richstyle I'd just suggest playing around with it until it accepts it?

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