# Homework Help: An implicit problem

1. Dec 9, 2011

### Twinflower

1. The problem statement, all variables and given/known data

Derivate $y$ with regards to $x, \frac{dy}{dx}$

$$2xy+y^2-4x=10$$

2. Relevant equations

3. The attempt at a solution

I am not very good at differentiation, so I havent got off to a good start, really.
I guess I can differentiate $y^2 and -4x$ "the usual way", but I am kind of stuck with the $2xy$-part.

I also recon I have to put the equation = 0

2. Dec 9, 2011

### SammyS

Staff Emeritus
To find the derivative of 2xy, use the product rule.

3. Dec 9, 2011

### I like Serena

Hi Twinflower!

How did you differentiate y2?

4. Dec 9, 2011

### Twinflower

So $(2x)'(y) + (2x)(y)' ---> y + 2x$ ?

That would yeld someting like: y + 2x +2y - 1, but the answer is supposed to be
$$\frac{dy}{dx} = \frac{2-y}{x+y}$$

5. Dec 9, 2011

### Twinflower

$(y^2)' = 2y$

I think :)

6. Dec 9, 2011

### SammyS

Staff Emeritus
(2x)' = 2, because x' = 1.

Where did y' disappear to?

7. Dec 9, 2011

### I like Serena

I'm afraid you can't simplify (2x)(y)'.
y' is what you want to find.

Not quite.
You should apply the chain rule here, and leave the y'.

8. Dec 9, 2011

### Staff: Mentor

No.
d/dx(y2) = 2y$\cdot$dy/dx = 2y$\cdot$y'

You didn't use the chain rule.

9. Dec 9, 2011

### Twinflower

Ok guys, I am gonna sit down and try to work this out.
Thanks for pushing me in the right direction.

I'll be back later :)

10. Dec 9, 2011

### Twinflower

OK, I finally got it:
$$2xy+y^2-4x=10$$

Differentiating each term:

$$\frac{dy}{dx}2xy = (2x)' \times y + 2x \times y' = 2y +2x \frac{dy}{dx}$$
$$\frac{dy}{dx}y^2 = 2y \frac{dy}{dx}$$
$$\frac{dy}{dx}-4x = -4$$
$$\frac{dy}{dx} 10 = 0$$

Combining the terms yelds:

$$2y +2x \frac{dy}{dx}+2y \frac{dy}{dx}-4=0$$

Diving each term by 2 and combining both y'-terms:

$$\frac{dy}{dx}\times (x+y) + y-2 = 0$$

Subtracting (y-2) on both sides:

$$\frac{dy}{dx}\times (x+y)=2-y$$

Dividing by (x+y) yelds the final answer:

$$\frac{dx}{dy}=\frac{2-y}{x+y}$$

Edit: Alcubierre posted the answer while I was creating mine :)

Last edited: Dec 9, 2011
11. Dec 9, 2011

### Alcubierre

Haha we posted pretty much at the same time. Congratulations though, good job!

12. Dec 9, 2011

### I like Serena

Good!

Btw, you did make a mistake here:
$$\frac{dy}{dx}y^2 = 2y + \frac{dy}{dx}$$
which should be:
$$\frac{dy}{dx}y^2 = 2y \cdot \frac{dy}{dx}$$
But apparently you fixed that while moving along.

13. Dec 9, 2011

### Twinflower

Ah, thanks Serena. Just a typo, it's not in my book.
Gonna fix my post to avoid any confusion in case others are trying to use this thread :)

14. Dec 9, 2011

### I like Serena

Hey Twinflower.

You still got 1 typo left, which is the one I actually mentioned. ;)

15. Dec 9, 2011

Done!