Question

Implicit differentiation:
y + 3xy -4 = 0

Answer 1

i see you lol

Answer 2

take derivative of each term as you would normally, except whenever you differentiate y multiply by y'

Answer 3

y'+3y+3xy'=0
y'+3xy'=4
y'(1+3x)=4
y'=4/(1+3x)

Answer 4

y'+3y+3xy'=0
y'+3xy'=4 <-- **4 should be -3y **
y'(1+3x)=4 <--
y'=4/(1+3x) <--

Answer 5

Thank you, that is what my book says as well. I'm a little bit lost though. I get that the derivative of the y = 1, multiply that by y\[\prime\], but then 3xy, why does that become 3xy[\prime\]? And the -4, how does that become 3y[\prime\]?

Answer 6

Oh, am I to treat it as (3xy)\[^{1}\]?

Answer 7

Ah sorry, my equation formatting is a mess :\

Answer 8

its the product rule
(fg)' = fg' +f'g

Answer 9

(3xy)' = (3x)*y' + (3x)'*y = 3xy' + 3y

Answer 10

I see. So then the 4 still becomes zero, as in other cases?

Answer 11

right

Answer 12

Okay, thank you very much :D That will be very helpful for the next problems, hopefully I can manage them on my own.

Answer 13

yw :)

Answer 14

Just double checking before I continue, for example: 3*x^3*y^2 would become 3x^3(2y) + 3(3x^2)(y^2), correct?

Answer 15

yes but don;t forget about the y'

Answer 16

So, each of those two terms is also multiplied by y prime? I'm not to clear on what you mean.

Answer 17

no only the term where you differentiated y (1st term)
3x^3(2y)y' because 2y came from differentiating y^2

Answer 18

Ah okay. So, I multiply by y(prime) whenever I differentiate y.