Question

get dy/ dx if this please.

x^3 - 7x^2y^3 + 4y^2 = -16

Answer 1

This is implicit differnetiation, so what we do is differntiatet both sides of the equation

Answer 2

so, we get this: 3x^2-14x2y^3+6y^2y'7x^2=0

Answer 3

Awww what the hey, I decided that if I sleep early I'll probably stay awake all night. If you want, I can show you how this is done. Wanna work on it together?

Answer 4

ok. let's do it.

Answer 5

Awesome. First thing we do is put each side in parentheses.

Answer 6

(x^3 - 7x^2y^3 + 4y^2) = (-16)

Answer 7

Then we write dy/dx next to each parentheses.
dy/dx(x^3 - 7x^2y^3 + 4y^2) = dy/dx(-16)

So far so good?

Answer 8

yeah,

Answer 9

Alright. Do you remember the rule for how you can differentiate each term separately if there's a plus or minus sign between them?

Like how dy/dx (4x^2 + 3x) can be split into dy/dx(4x^2) + dy/dx(3x)

Answer 10

(differentiate is what some people accidentally call "deriving", just in case you were wondering)

Answer 11

Lemme know if that part makes sense so far, otherwise it'll be too confusing if we start the next steps.

Answer 12

yeah, i'm still getting it.

Answer 13

Awesome. So go ahead and tell me how we will rewrite the current equation,
dy/dx(x^3 - 7x^2y^3 + 4y^2) = dy/dx(-16)
when we separate it by parts into separate dy/dx parts.

Answer 14

(Don't do any actual math on it yet. Just distribute the dy/dx)

Answer 15

Again, if anything's confusing, please let me know before you bail out. For some reason lots of folks bail out instead of telling me that they're stuck. :P

Answer 16

Did you give up? :(

Answer 17

hey i'm still on it.

Answer 18

Ah ok. I thought you left me. :P
Remember: you just have to set it up, don't do any math yet.

Answer 19

yes, i know

Answer 20

Okie dokie. Just post it here when you've set it up.

Answer 21

this is my answer. y' = (3x^2) / [2y (21x^2 - 4)]

Answer 22

Okie dokie, lemme check it out. You did the math instead of just setting it up, so I wasn't quite ready to get an answer. Haha.

Answer 23

sorry. i'm quite on a hurry.

Answer 24

y' = (3x^2-14xy^3)/(7x^23y^2+8y)
is what I got.

Answer 25

i answered it wrong! i'm so sorry.

Answer 26

let me try again please.

Answer 27

Not a problem. Sure thing. But let's do it step by step. Just tell me what it looks like after this step: dy/dx(x^3 - 7x^2y^3 + 4y^2) = dy/dx(-16)

Just show me where we put each dy/dx after we get rid of the ( )

Answer 28

y' = (3x^2 - 14xy^3) / (21x^2y^2 - 8y)

this is my answer. please kindly check it. :)

Answer 29

That isn't what I got. :/ But let's just go through it one step at a time, that way we can correct each other if one of us makes a mistake. I'll go ahead and do the step that I've been begging you to try.

dy/dx(x^3 - 7x^2y^3 + 4y^2) = dy/dx(-16)
equals
dy/dx(x^3) - dy/dx(7x^2y^3) + dy/dx(4y^2) = dy/dx(-16)

Does this make sense so far, and does it look correct?

Answer 30

geez. i really don't have any more time. my exam will be in about 20 minutes and i still have to go to school. i'm so sorry. but you're a good teacher so far. i hope you can teach me again next time. just we me luck. :)

Answer 31

Sure thing. Good luck. Just one quick notice:
dy/dx(7x^2y^3)
On this step MAKE SURE YOU USE THE PRODUCT RULE! Just in case you haven't.
And remember: dy/dx of y^2 is 2y*y'

Answer 32

yep. i'll keep it in mind.. lots of thanks!

Answer 33

and constants are automatic zero! :))

Answer 34

Awesome! I think you'll do very well. Get back to me after the test if you have any questions!

Answer 35

Kinda. Just one mistake: does 1 + 2 = 31?