# Homework Help: Trying to figure out how derivatives work.

1. Sep 8, 2012

### vysero

If a particle's position is given by x = 4-12t+3t^2 (where t is in seconds and x is in meters).

a) What is the velocity at t = 1s?

Ok, so I have an answer:

v = dx/dt = -12 + 6t
At t = 1, v = -12 + 6(1) = -6 m/s

but my problem is I want to see the steps of using the formula v = dx/dt in order to achieve -12 + 6t...

I am in phx with calc and calc is only a co-requisite for this class so im taking it while im taking physics but as you can see calc is a little behind. Where just now learning limits in calc and I was hoping someone could help me figure this out.

2. Sep 8, 2012

### cepheid

Staff Emeritus
There is a differentiation rule called the power rule for power law functions which states that $$\frac{d}{dx}(x^n) = nx^{n-1}$$ You should be able to prove this using the definition of a derivative (in terms of limits) that you have learned. Another rule is that the derivative of a sum of functions is equal to the sum of the derivatives of those individual functions (which is also easy to prove). So you can differentiate each term in your expression separately, and add the results. The only difference is that here your independent variable is time (t) rather than x.

3. Sep 8, 2012

### vysero

Sorry I am not catching on. Ok lets say I know how to find the derivative of f(x) = x^2 (which I just learned to do after watching like 6 video's on khan's academy) if I find the derivative of x = 4-12t+3t^2 (which is odd to me because it seems like I would need to say that f(y) = 4-12t+3t^2 in order to do that) would that get me anywhere?

4. Sep 8, 2012

### cepheid

Staff Emeritus
I'm sorry that it was so arduous. But all you have to do is use the equation I gave above. In your example, n = 2.

Be careful. I think that you are confusing what is the dependent variable, and what is the independent variable. In a general mathematics context, we often denote the dependent variable by 'y' and the independent variable by 'x'. The variable 'y' is "dependent" because it is a function of x: y = f(x).

HOWEVER, in this physics context, we're using different symbols for our variables. The independent variable is time, so we use the symbol 't' to denote that. The dependent variable is horizontal position, so we use 'x' to denote that ('x' as in the x-coordinate). There is no y here. So, in this case, position x is a function of time t:

x = f(t)

And you want to find the derivative of the function f(t): $$\frac{dx}{dt} = \frac{d}{dt}(4 - 12t + 3t^2)$$To do this, you will need to know two facts:

1. The power rule, as I described above
2. The fact that the derivative of a sum of functions is equal to the sum of the derivatives of those functions. This means that you can consider each term in the expression to be a separate function, you can differentiate each of these functions separately, and then you can add the results together.

5. Sep 8, 2012

### vysero

So the d in dx/dt does not stand for derivative does it, it stands for differentiate?

Last edited by a moderator: Sep 8, 2012
6. Sep 8, 2012

### rcgldr

Link to a web page showing how find the derivative of x2 via a limit, shown in answer 1. On that web page, h is used instead of Δt.

http://www.math.montana.edu/frankw/ccp/calculus/deriv/derlimit/learn.htm [Broken]

Last edited by a moderator: May 6, 2017
7. Sep 8, 2012

### cepheid

Staff Emeritus
Ohh boy, okay. We need to backtrack a little. Let's talk a bit about terminology and notation:

1. To "differentiate" a function means to take the derivative of that function. So we say "differentiate", we DON'T say "derive" or "derivate." So the action of computing a derivative is called "differentiating." Similarly, "differentiation" is the noun for that action or process.

2. There are several different ways to denote a derivative. I.e. there are a bunch of different mathematical notations.

a. The notation of Newton, which is to use a prime symbol ($\prime$) next to the function, to indicate that it has been differentiated once. So, if you have a function $f(t)$, then the first derivative of that function with respect to t would be denoted as $f^\prime(t)$. You add another prime symbol for the second derivative: $f^{\prime\prime}(t)$. Note that in physics, we often don't bother introducing a second letter for the function notation. So, in your example, instead of saying $x = f(t)$, we would just write the function as $x(t)$, which indicates in a compact way that x is a function of t. Then the derivative would be $x^\prime(t)$.

b. The notation of Leibnitz, which is as follows: if we have a function x = f(t), then the derivative of that function with respect to time is given by the symbol:$$\frac{df(t)}{dt}$$which can also be written as$$\frac{d}{dt}f(t)$$ The "d" just by itself doesn't mean anything. The entire expression $\frac{d}{dt}$ should be regarded as ONE symbol that acts on the function. (It's NOT a fraction or a division of two things) In fact, when we have something that acts on a function to carry out some sort of mathematical operation on it, we call that an operator. So the symbol $\frac{d}{dt}$ can be regarded as the "differential operator". In other words, it is the operator that carries out the operation of differentiation with respect to time. Any function that you put next to this operator is being differentiated with respect to time. That's why the second way of writing it: $\frac{d}{dt} f(t)$ is more suggestive, because it makes it clearer that the differential operator d/dt is acting on the function f(t) to produce the derivative of f(t) with respect to time.

These two notations MEAN the same thing. Both of them are ways of writing a derivative. I.e.$$f^\prime(t) = \frac{df(t)}{dt}$$

As I mentioned before, physicists are lazy/sloppy with notation. So, OFTEN in the Leibnitz notation, the argument of the function is omitted altogether, and we just write $\frac{df}{dt}$. Furthermore, just like I said above for the Newton notation, we sometimes don't bother with the 'f' at all, and just write the function using same symbol as its variable: x(t). In this case, for the Leibnitz notation, the derivative would be written as $\frac{dx}{dt}$. In fact, with the Leibnitz notation, the letter for the dependent variable (y, or in this case x) is what appears more often than the letter for the function (f or g or whatever), assuming the two are even distinct. I.e. if y = f(x), then in the Leibnitz notation, you'd might more often write dy/dx, rather than df/dx, for the derivative.

Again, I want to emphasize that although dx/dt (or df/dt) looks like a ratio of two quantities, it is not. It's just a symbol for "derivative of x with respect to t." So why did Leibnitz choose to write it that way? To be honest I don't know, but to me, I think it is meant to be suggestive. For, although dx/dt is NOT a ratio, the way you compute is by causing a small change Δt, and seeing what the resulting change Δx is. The ratio of these two Δx/Δt is the average rate of change of the function x(t) over the interval Δt. To figure out the instantaneous rate of change of x, i.e. how fast x is changing at particular instant in time, you take the LIMIT of this ratio Δx/Δt as Δt → 0 (as you are no doubt learning in your math class). This *limit* is the formal definition of the derivative of the function: dx/dt. So although dx/dt is not a ratio, its value is equal to the limit of the ratio Δx/Δt.

Again physicists are a bit sloppy and DO tend to think of dx/dt as a ratio of an "infinitesimal" (i.e. vanishingly small) change in position dx to an "infinitesimal" change in time dt (as opposed to Δx and Δt, which were finite changes in those quantities) So they DO treat dx/dt as a fraction, and sometimes even separate the dx and the dt, even though this is not strictly correct and is an "abuse" of notation. But I mention it because you will no doubt see it in your physics classes. The reason why it is frowned upon somewhat is because, at least at the intro calculus level, we have no precise way of defining what an "infinitesimally small" quantity is. So this is not a well-defined concept. Instead, we define the derivative using the concept of a limit, which IS a precisely-defined concept in mathematics.

Last edited: Sep 8, 2012
8. Sep 8, 2012

### vysero

First of all I want to say thank you for taking so much time to explain this stuff to me. I really do appreciate all of your help.

Let me see so d/dx(f(x)) means I am taking the derivative of f(x).

So since d/dx(x^n)=nx^n-1 then taking d/dx(x^n) is a shorter way of coming up with a derivative for (x^n)?

So then -12 + 6t is actually the derivative of 4-12t+3t^2? Or am I still not getting it?

9. Sep 8, 2012

### cmmcnamara

You are on the right track, definetly!

d/dx(f(x) does mean that you are taking the derivative of f(x) with respect to x.

d/dx(x^n)=nx^(n-1) is the derivative of any function of the form x^n. Essentially it is the rule for the derivative of a function of that form. It allows us to immediately write the derivative for that function with going through all the steps required by the definition of derivative. So yes, it is a short way to get the derivative of x^n. You should note however that n cannot be equal to zero. If you want to read up on it more in depth: http://en.wikipedia.org/wiki/Power_rule

Yes -12+6t is the derivative. This is by applying the power rule and the derivative's linearity properties by adding each derivative of each function.

10. Sep 8, 2012

### cepheid

Staff Emeritus
Yeah

d/dx(xn) IS the derivative of xn. But how do you evaluate (or compute it)? Well, you could use the definition of a derivative, and take the limit. But this is tedious. A shorter way *to compute it* is to use the power rule for differentiation. The power rule says that $\frac{d}{dx}(x^n) = nx^{n-1}$ for any n. You can prove once that the power rule is true using the limit definition of a derivative, and from then on, just go ahead and use it.

Yes.

To see this: remember that I said that you can differentiate each term separately? Well d(4)/dx = 0, because the derivative of a constant function is 0. (The derivative of a function is the rate of change of that function. If the function is constant, then it doesn't change, so its rate of change is 0). What about the other two terms? You can just use the power rule on each of them, and you should get the answer above.

11. Sep 8, 2012

### vysero

I get it sweet!!!! Thank you guys so much!!! I wish I could send you a card or something lol thanks a lot! I am like guna be ahead of everyone in my math class for like the first time in my life!!! haha

12. Sep 8, 2012

### cepheid

Staff Emeritus
You're welcome! It's pretty common for your physics profs to expect you to know math that you haven't been formally taught yet (and not just in first year). I've been there, and I know how frustrating it can be. That's why I put some extra effort into explaining the basics.