anonymous
  • anonymous
Implicit differentiation: y + 3xy -4 = 0
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
i see you lol
dumbcow
  • dumbcow
take derivative of each term as you would normally, except whenever you differentiate y multiply by y'
lalaly
  • lalaly
y'+3y+3xy'=0 y'+3xy'=4 y'(1+3x)=4 y'=4/(1+3x)

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dumbcow
  • dumbcow
y'+3y+3xy'=0 y'+3xy'=4 <-- **4 should be -3y ** y'(1+3x)=4 <-- y'=4/(1+3x) <--
anonymous
  • anonymous
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\]?
anonymous
  • anonymous
Oh, am I to treat it as (3xy)\[^{1}\]?
anonymous
  • anonymous
Ah sorry, my equation formatting is a mess :\
dumbcow
  • dumbcow
its the product rule (fg)' = fg' +f'g
dumbcow
  • dumbcow
(3xy)' = (3x)*y' + (3x)'*y = 3xy' + 3y
anonymous
  • anonymous
I see. So then the 4 still becomes zero, as in other cases?
dumbcow
  • dumbcow
right
anonymous
  • anonymous
Okay, thank you very much :D That will be very helpful for the next problems, hopefully I can manage them on my own.
dumbcow
  • dumbcow
yw :)
anonymous
  • anonymous
Just double checking before I continue, for example: 3*x^3*y^2 would become 3x^3(2y) + 3(3x^2)(y^2), correct?
dumbcow
  • dumbcow
yes but don;t forget about the y'
anonymous
  • anonymous
So, each of those two terms is also multiplied by y prime? I'm not to clear on what you mean.
dumbcow
  • dumbcow
no only the term where you differentiated y (1st term) 3x^3(2y)y' because 2y came from differentiating y^2
anonymous
  • anonymous
Ah okay. So, I multiply by y(prime) whenever I differentiate y.

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