Basic derivative notation question

**Phire** · May 29th 2009, 02:06 PM

I understand P'(t) alot easier, but I'd like to understand the other notation. (p = position, t = time, v = velocity, a = acceleration).

$\frac{dp(t)}{dt}$

My first question.. is this an actual fraction? What's in the numerator, and what's in the denominator?

Is the input variable necessary in the top? Could it just be:

$\frac{dp}{dt}$

Now, about stacking derivatives:

$\frac{d^2p(t)}{dt^2}$

Ok.. are these actual exponents or are they just notations to indicate how many derivatives have been done? And why is it that the "exponent" is placed on d in the numerator and placed on t in the denominator?

**craig** · May 29th 2009, 02:36 PM

This particular method of writing derivatives is Leibniz's notation, there is some interesting information of information on Wikipedia about this type of notation.

Hope this helps

**HallsofIvy** · May 30th 2009, 06:18 AM

Originally Posted by Phire

I understand P'(t) alot easier, but I'd like to understand the other notation. (p = position, t = time, v = velocity, a = acceleration).

$\frac{dp(t)}{dt}$

My first question.. is this an actual fraction? What's in the numerator, and what's in the denominator?

No, it is not an "actual fraction"- it is defined exactly like p'(t) is: $\lim_{h\to 0}\frac{p(t+h)- p(t)}{h}$ . However, it has a nice property: you can go back before the limit, use the fraction property, and then take the limit so all "fraction properties" hold! You can define "differentials": "dx" is simply symbolic and we define "dp= p' dx" to make use of the fact that, although dp/dx= p' is NOT a fraction it can be treated like one! That's a very "vague" definition. "Differentials" are defined more precisely and used a lot in higher mathematics like Differential Geometry.

Is the input variable necessary in the top? Could it just be:

$\frac{dp}{dt}$

Yes, if it is understood what the variable is. That's commonly done. And, of course, we first define the derivative at a specific point, with the derivative function being defined as the function that takes on those values at each point. If I want to talk about the derivative of p(t) at t= 3, I would have to write $\frac{dp(3)}{dt}$

Now, about stacking derivatives:

$\frac{d^2p(t)}{dt^2}$

Ok.. are these actual exponents or are they just notations to indicate how many derivatives have been done? And why is it that the "exponent" is placed on d in the numerator and placed on t in the denominator?

No, those are not exponents. It is also true that higher derivatives like that [b]cannot[\b] be treated as fractions. The position of those numbers is just custom. One reason, perhaps, is to remind us that they CANNOT be cannot be treated like a fraction.

**craig** · May 30th 2009, 06:36 AM

Originally Posted by Phire

Ok.. are these actual exponents or are they just notations to indicate how many derivatives have been done? And why is it that the "exponent" is placed on d in the numerator and placed on t in the denominator?

As HallsofIvy said these are not exponentials, and the reason the $^2$ is placed differently could be something to do with the fact that the second derivative is:

$\frac{d}{dx}[\frac{dy(x)}{dx}]$

Which is pretty much equal to:

$\frac{d^{2}y(x)}{dx^{2}}$

Probably a combination of this and for clarity's sake.

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