# Calculating dy

1. Aug 28, 2011

### vanmaiden

1. The problem statement, all variables and given/known data
I was messing around online when I found this: $\frac{dy}{2}$ = 2x. This was derived from the function y = x2. I had never really seen anything like this before. When I solved for "dy," I got 4x. However, for example, when x changes from 0 to 2, the y changes from 0 to 4. Interestingly enough, "dy" is represented as being 4x and not just 4. Can someone point out what I did wrong?

2. Relevant equations
y = x2
$\frac{dy}{2}$ = 2x

3. The attempt at a solution
Pretty much explained in the "problem statement" by accident.

2. Aug 28, 2011

### SteamKing

Staff Emeritus
You find a lot of things on the internet. Shockingly, not all of them are true.

3. Aug 28, 2011

### vanmaiden

That is true without a doubt! lol. How would I go about solving something like this though? I am given delta x and the derivative and have to find delta y. It all seems very interesting.

4. Aug 28, 2011

### HallsofIvy

Staff Emeritus
You wouldn't. The equation is meaningless. You cannot have a "dy" without a corresponding "dx". In terms of "non-standard analysis", we would say that the left side of the equation is an infinitesmal while the right side is not. That can't happen.

Since you say "This was derived from the function $y = x^2$" I suspect that it was supposed to be
$$\frac{dy}{dx}= x^2$$

5. Aug 28, 2011

### Staff: Mentor

The equation above should be dy = 2x dx

6. Aug 31, 2011

### stallionx

Here the point is, in my personal opinion

dy = 4*x

x is a variable but as the problem ( lamely presented ) offers

dy=0

and

thus
x=4*0

x=0 alwyas no matter what

Either wrongly presented or gives us the line equation

x = 0

7. Aug 31, 2011

### vanmaiden

Thank you

8. Aug 31, 2011

### stallionx

You are quite Welcome :)

9. Aug 31, 2011

### Staff: Mentor

As HallsOfIvy already pointed out, this equation is meaningless, so there is not much point in analyzing it further.