Question

Could someone help me with this derivative?

Answer 1

How did they get this?

Answer 2

looks like you got it right

Answer 3

oh, the question was HOW?\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\]

Answer 4

I did, but I was just inputting answers that I got from a classmate (which we're allowed to do). I just don't know how they got it. I tried the quotient rule, but I didn't get the -y(y-12).

Answer 5

in this case \[f(y) = y-6, f'(y) = 1, g(y) = y^2+2y+12, g')y) = 2y+2\]

Answer 6

the numerator is \[y^2+2y+12-(y-6)(2y+2)\]

Answer 7

g(y) is y^2 - 2y + 12

Answer 8

i guess algebra does the rest \[y^2-2y+12-(2y^2+2y-12y-12)=...\]

Answer 9

you good from there, or need more steps
it is all algebra from here on in
combine like terms etc

Answer 10

ok i made a typo

Answer 11

I still didn't get it, I ended up with this again: I have y^2 - 2y + 12 - (2y^2 - 2y - 12y + 12) -> -3y^2+ 12y ->

Answer 12

\[y^2-2y+12-(y-6)(2y-2)\]

Answer 13

ok lets go slow

Answer 14

I have y^2 - 2y + 12 - (2y^2 - 2y - 12y + 12) ->
y^2 - 2y + 12 - 2y^2 + 2y + 12y - 12
, right so far?

Answer 15

\[y^2-2y+12-(y^2-12y-2y+12)\]

Answer 16

yeah so far so good

Answer 17

y^2 - 2y + 12 - 2y^2 + 2y + 12y - 12
-2y and 2y cancel. +12 and - 12 cancel. and what's left is y^2 - 2y^2 + 12y right?

Answer 18

keep going

Answer 19

Ahh. I was subtracting wrong. Then it's -y^2 + 12y which is -y(y - 12). There we go. I kept subtracting the y^2 - 2y^2 wrong.

Answer 20

here is a true fact

Answer 21

i can always make elementary algebra students get a question incorrect if i they have to combine like terms with \(x-2x\) or say \(x^2-2x^2\)
one minus two screws them all up every time!

Answer 22

but yeah, you got it
\[5x-12x=-7x\] is never a problem however, not sure why

Answer 23

No idea why! I kept seeing it as -y^2 - 2y^2 which is -3y^2. Just had to slow down for a bit. I'm trying to race against time to study enough for my exam. Thank you!

Answer 24

yw, good luck with your exam

Answer 25

Thanks!