Question

Find dy/dx by implicit differentiation.

x^2 + xy + y^2 = 4.

Okay, so I know that I have to differentiate the problem so that x^2 becomes 2x and y^2 becomes 2y • dy/dx and 4 becomes 0. What should I do with xy?
I know I need to use the product rule, but my answer just isn't matching up to the answer I was given.

Answer 1

d/dx uv = u'v + uv'
where v' would be y' or dy/dx

Answer 2

Okay, so it would be x'y + xy' which would then equal y + x, right?

Answer 3

y+xy'

Answer 4

Okay, so it would be x + y times dy/dx, right?

Answer 5

how'd you get that?

Answer 6

Because y' = dy/dx and the problem wants us to work with dy/dx.

Answer 7

this is what is happening, you are differentiating term by term (each term separately).
You are differentiation X's and Y's all the same y, BUT whenever you are differentiating Y you are multiplying that times dy/dx (or times y', -- whatever notation satisfies you)

as you differentiate all terms, solve for y' (or for dy/dx -- whatever you wrong the derivative as)

Answer 8

my point is \(\dfrac{d}{dx}xy=x'y+xy'=y+x\dfrac{dy}{dx}\)
you flipped the x and y

Answer 9

Ohhh, oops! I didn't mean to do that, I'm sorry!

Answer 10

don't be sorry, just don't do it again

Answer 11

jk

Answer 12

"wrong" in last parenthesis shouldn't be there "it is supposed to be "write"

Answer 13

@SolomonZelman Wait, what do you mean by your last statement?

Answer 14

I am correcting my first one

Answer 15

he typed wrong instead of wrote

Answer 16

Ohhh, okay

Answer 17

yeah

Answer 18

Okay, so I'm at the point where I have this: 2x + (y + x•dy/dx) + 2y = 0. Is this correct so far?

Answer 19

no

Answer 20

the derivative of y^2 is also multiplied times y'

Answer 21

Dang it, you're right. So it would be 2y times dy/dx?

Answer 22

yes

Answer 23

Okay, so I have 2x + (y + x•dy/dx) + 2y•dy/dx= 0. I know I need to get dy/dx alone on one side. I can subtract 2x at least, but what do I do about the rest?

Answer 24

\(\large\color{black}{ \displaystyle 2x+y+\frac{dy}{dx}x + \frac{dy}{dx}2 y = 0 }\)

(you can use my code,
```
\(\large\color{black}{ \displaystyle 2x+y+\frac{dy}{dx}x + \frac{dy}{dx}2 y = 0 }\)
```
)

Answer 25

yes

Answer 26

now isolate \(\large\color{black}{ \displaystyle \frac{dy}{dx} }\)

Answer 27

Okay, so I've got this so far: x•dy/dx + 2y•dy/dx = =2x - y. I can divide to get 2y and x on the other side. I'm a bit confused at this part.

Answer 28

-2x ? on the right

Answer 29

you are subtracting 2x from both sides aren't you?

Answer 30

-2x, my bad!

Answer 31

yes, then factor out of dy/dx on the left side

Answer 32

That was a typo on my part.

Answer 33

it's fine

Answer 34

Okay, but if I divide by both x and 2y to isolate dy/dx, won't it become dy/dx + dy/dx? Or am I missing something?

Answer 35

I would say messing something, because I don't get what you mean...

\(\large\color{black}{ \displaystyle 2x+y+\frac{dy}{dx}x + \frac{dy}{dx}2 y = 0 }\)

\(\large\color{black}{ \displaystyle \frac{dy}{dx}x + \frac{dy}{dx}2 y = -2x-y }\)

\(\large\color{black}{ \displaystyle \frac{dy}{dx}(x + 2 y )= -2x-y }\)

and then the last step

Answer 36

Ooooohhhhhh! I get it now!

Answer 37

So then it would become dy/dx = -2x - y / x + 2y!

Answer 38

yes, but don't put the exclamation mark

Answer 39

and put the parenthesis

Answer 40

Aha, sorry about that. However I do have one more question. The answer for this problem that I was given is (2x + y)/(2y - x). How come these answers don't match up?

Answer 41

it is wrong

Answer 42

My answer or the answer I was given?

Answer 43

we did all steps together. Our answer is fine

Answer 44

the answer you were given intially is wrong

Answer 45

Ahhh, okay. Thank you all who helped me so much, I really appreciate it.

Answer 46

Confirmation
http://www.wolframalpha.com/input/?i=derivative+x%5E2+%2B+xy+%2B+y%5E2+%3D+4.

Answer 47

yes, our answer is right. Wolfram also agrees to us.

Answer 48

Okay. Thank you all again so much!

Answer 49

Sure!