Question

Implicit Differentiation: find dy/dx for:
x^2 +y^2 +3xy =9
Help appreciated!

Answer 1

I am a bit confused with these types of problems.
Here's what I've got:
2x(dx) +2y(dy) +3x(dy) +3y(dx)
... and I'm stuck. Have I done this correctly? What should I do next?

Answer 2

http://openstudy.com/users/loser66#/updates/52b47e8fe4b03e81ff95fad9

Answer 3

exactly yours.

Answer 4

ok, cool! :) I'll go check that out then, thank you!

Answer 5

hm, the steps are kind of hard to read with the drawings. I see the answer is the same one I had gotten, though, but I'm still a bit iffy on the process. I'll keep looking at it.

Answer 6

ok, we can do it again, show me your work

Answer 7

ok well I know the derivative of a constant is 0 so I just have 0 on the right?
so on the left I would have x^2 which is 2x(dx) and y^2 which is 2y(dy) and 3xy which is 3x(dy)+3y(dx) because of Product Rule?
so I would have 2x(dx) +2y(dy) +3x(dy) +3y(dx) =0
good so far?

Answer 8

\[(x^2 +y^2+3xy)'= (x^2)' +(y^2)'+(3xy)'\] OK?

Answer 9

yes ok

Answer 10

you put dx and dy at the end of the term . It makes you confuse. take it off

Answer 11

\((x^2)' = 2x\)OK?

Answer 12

ok so just the apostrophe instead?

Answer 13

\((y^2)'= 2yy'\) OK?

Answer 14

ok because y^2 goes to 2y and y' means it is a derivative term we will be separating?

Answer 15

not that reason, the reason is y is a function respect to x. And x itself is a variable only.

Answer 16

ah, ok.

Answer 17

the third term is \((3xy)' =3\color{red}{x'}y+3x\color{blue} {y'}=3y +3xy'\) OK?

Answer 18

oh, I see! thanks for the color :)

Answer 19

and now just combine them and the right handside =0

Answer 20

2x +2yy' +3y +3xy' =0 ?

Answer 21

\[2x + 2y\color{red}{y'}+3y+3x\color{red}{y'}=0\]
factor y' and isolate it to get the answer

Answer 22

so to factor y' should I get terms without it to the other side?
2x+3y = -2yy' -3xy'
2x+3y= y'(-2y-3x)
2x+3y/-2y-3x = y' ?

Answer 23

yup

Answer 24

hm I was just looking at the two closest choices (this is multiple choice):
a) 3x+2y/-3y-2x
b) -3y-2x/3x+2y
b has to be the answer if I look closely, but it seems I just have the negatives in the wrong place? I think it is just that I might have distributed the -1 in the wrong place, because this is what I got the first time:
-(2x+3y/3x+2y)

Answer 25

hey hey, your is b

Answer 26

think of let - sign out of the denominator and put it into the numerator, it's b. They are the same, EXACTLY

Answer 27

so I guess i'm just double checking. hm, I know I did this correctly and b is the answer. I just have (2x+3y)/(-2y-3x) on my paper
but the answer is (-3y-2x)/(3x+2y)
so just a strangely distributed -1?

Answer 28

right, ok. it works either way.

Answer 29

hehehehe....

Answer 30

thanks so much for your help!