Question

Calculate the derivative of the function
s(x) =

3x − 9
2x + 4
3

Answer 1

\[thats (\frac{ 3x-9 }{ 2x+4 })^{3}\]

Answer 2

I got \[(\frac{ 3x-9 }{ 2x+4 })^{2}(\frac{ 9x-27 }{ (2x+4)^{2} })\] but thats incorrect, i cant find my mistake

Answer 3

\[\Large \left[\left(\frac{ 3x-9 }{ 2x+4 }\right)^{3}\right]'\quad=\quad 3\left(\frac{ 3x-9 }{ 2x+4 }\right)^{2}\color{royalblue}{\left(\frac{3x-9}{2x+4}\right)'}\]Hmm let's see what the quotient rule gives us.
I think you may have made a little boo boo somewhere.

Answer 4

\[\Large \color{royalblue}{=\quad\frac{(3x-9)'(2x+4)-(3x-9)(2x+4)'}{(2x+4)^2}}\]
\[\Large \color{royalblue}{=\quad\frac{(3)(2x+4)-(3x-9)(2)}{(2x+4)^2}}\]

Answer 5

@zepdrix oh so i combine the two? So the answer is what typed last?

Answer 6

i thought i used quotient rule on the right part and power rule for the left, then wrote both for the answer

Answer 7

This is the answer before quotient rule has been applied to the blue part:\[\Large 3\left(\frac{ 3x-9 }{ 2x+4 }\right)^{2}\color{royalblue}{\left(\frac{3x-9}{2x+4}\right)'}\]Then the blue part is expanded below, plug the blue back into the blue :)
It's kinda lengthy so I was trying to stay organized by not rewriting everything.
Yes the answer is both parts.

Your answer was very close, you shouldn't have gotten a `9x-27` in your numerator though.
I think it works out to `30` instead.

Answer 8

@zepdrix 30 for the numerator doesn't work either, thats still incorrect somehow

Answer 9

Hmm ok sec :x

Answer 10

Hmm :(

Answer 11

@zepdrix was 3 multiplied into 30 already??