# Derivative notation

1. Mar 15, 2010

### PhysicsMark

1. The problem statement, all variables and given/known data
Is it correct to say that if "y" is a function of x, then

$$y'=\frac{dy}{dx}$$

??

2. Mar 15, 2010

### mg0stisha

Yes (assuming x is your only variable).

3. Mar 15, 2010

### Staff: Mentor

Those are two different ways of writing one thing.

The y' notation is sometimes called Newton notation (although he used a dot instead of a prime), and the other is called Leibniz notation.

4. Mar 15, 2010

### PhysicsMark

I think I may have answered my own question with this next one, but I would like to get a confirmation.

I asked the question above because in my math methods in physics class, the tutorials we use and the professor often algebraically manipulate the dy's and dx's in equations involving y'.

I am reviewing scale invariance and its applications to FODE's.

After proving the following equation is scale invariant, I am to solve for the general solution.

$$y'+\frac{y^{2}}{x^{2}}=2$$

After I have shown it is scale invariant, I use the substitution y=vx to obtain:

$$v+x\frac{dv}{dx}+v^2=2$$

In order to get to the point above, I note that:

$$dy=vdx+xdv$$

Here is where my question comes in. I originally looked at this problem by stating:

$$y'=\frac{dy}{dx}=\frac{d(vx)}{dx}=\frac{dvdx}{dx}=dv$$

This is the wrong simplification

As far as I know, v is a number not a function of x. I believe I understand that this method is used to make an inhomogeneous equation separable. How do I avoid making the wrong simplification?

I think the answer is that I need to recognize that I must have a "dv" and a "dx" in order to solve the equation in a separable manner, otherwise I would have had:

$$dv+v^2=2$$

Does that make any sense at all? Thanks for replying.

5. Mar 15, 2010

### PhysicsMark

Haha...nevermind. It is the same thing, just not carried out all the way through.

dy/dx=dv........dy=dv(dx)= vdx+xdv.