# Derivatives of implicit functions

1. Oct 7, 2005

If you are given the derivative of an implicit function as $$y' = \frac{y}{2y+x}$$ how would you find all points (x,y) such that the slope at those points is 1/2? Ok so I did: $$\frac{y}{2y+x} = \frac{1}{2}$$ and got x = 0. So if I substitute x = 0 back into the original equation I get $$(0, \sqrt{2})$$. Would this be the correct method to solve this question? Also show that the slope is never equaled to 0. So I did $$\frac{y}{2y+x} = 0$$. If y = 0, then the original equation wouldnt make sense. Is this a valid response?

Any help is appreciated

Thanks

2. Oct 7, 2005

### Jameson

Well, I wouldn't word it "the original equation wouldn't make sense". Can you word this more mathematically or demonstrate why cannot equal 0?

3. Oct 7, 2005

### amcavoy

When you solve:

$$\frac{y}{2y+x}=\frac{1}{2}\quad (1)$$

and come up with x=0, I don't see how you get the square root of 2. Try plugging x=0 into equation (1). The y variable will cancel out and you get an identity. What does that tell you about the possible values of y?

Alex

4. Oct 7, 2005

### Jameson

He gave y', so I thought he plugged 0 into y, which he didn't give in his post.

5. Oct 7, 2005

### mezarashi

I think you forgot to note that equation one is the derivative of the function. I think plugpoint may have forgot to post the original equation to where he got the square root 2. The current differential is not separable.

6. Oct 7, 2005

### amcavoy

You don't need the original equation. If you plug x=0 into y', you get 1/2 no matter what value of y you have. This means that along the vertical line x=0 the slope is the same for all values of y.

Alex