## anonymous 4 years ago Implicit differentiation: y + 3xy -4 = 0

1. anonymous

i see you lol

2. dumbcow

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

3. lalaly

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

4. dumbcow

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

5. 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\]?

6. anonymous

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

7. anonymous

Ah sorry, my equation formatting is a mess :\

8. dumbcow

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

9. dumbcow

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

10. anonymous

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

11. dumbcow

right

12. anonymous

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

13. dumbcow

yw :)

14. 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?

15. dumbcow

yes but don;t forget about the y'

16. anonymous

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

17. dumbcow

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

18. anonymous

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