# Can someone tell me the essence of derivative?

1. Jan 28, 2004

### franz32

How can I "easily" solve or understand the application of derivative involving the rate of change?

2. Jan 28, 2004

### matt grime

This is a bit tricky to answer because the notion of 'rate of change' is to me the easy explanation of what the derivative is.

Differentiation from first principles.

Suppose we want to know the slope of the graph of some function at a point. We can try drawing the tangent line by hand and hoping we get a good enough fit and then finding its gradient (it's just a straight line after all). But the problem there is the standard 'real life does not correspond to the perfect world we imagine mathematically'.

Instead let's think about drawing a little chord from the point on the curveto a little point a bit further along, say e in the x direction. The smaller we let e get the better that chord's slope approximates the tangent's.

In numbers, if f is the function, we want to know what happens in

{f(x+e)-f(x)}/e

as e gets small.

To see where that expression comes from, try drawing the graph of something, and picking some x, some e and 'joining up the dots... '

I'm sorry, anyone know if the tex mode here allows us to use xypic?

In words, that quantity is looking at the instantaneous rate of change if you like. We'll try and explain why later.

Let's do an example

x^n

Work out the binomial expanion of (x+e)^n and subtract x^n, divide everythin by e and what happens as e gets small? As is often the case, here you can just set e to be zero and nothing goes wrong. You should get nx^(n-1).

Do the same for sin, knowing that for e small sin(e) is practically e.

Let's try and get back to rates of change.

Kinetics. Suppose we are doing th standard eqautions of motion with a fixed acceleration.

suppose our initial velocity is u, we accelerate for e seconds what's our final velocity? u+ae where a is the acceleration.

Now, a is the rate of change of speed with respect to time, as we all know and understand. What this tells us is, writing it more formally,

let v(t) be the velocity at time t

v(t+e)=v(t) + e.a

where a can be thought of as dv/dt

generally, dv/dt isn't constant, and the above equation should be approximate. So another way to interpret (actually just the same but rewritten) derivatives is

f'(x) is the function that makes

f(x+e) apporximately equal to f(x) + e.f'(x)

it's the linearization of the error if you like. The approximate we needn't worry about too much at the moment.

How much of that do you know?

What do you mean by 'solve' an application? Perhaps if you posted an example of what you were trying to solve?

Last edited: Jan 28, 2004
3. Jan 28, 2004

### franz32

Me again...

Well, I was wondering how can I find the derivative of an area (let's say circle) with respect to its radius... sort of those kinds of problem solving.

Is there a general explantion for these kinds of problem?

4. Jan 28, 2004

### Tom Mattson

Staff Emeritus
Re: Me again...

Yes, and that's what Matt was driving at. When he says:

He means that you take the limit of that expression as e-->0. In your case, you have A=f(r)=pr2.

So, take [f(r+e)-f(r)]/e and take the limit as e-->0.

Like so:

f'(r)=lime-->0[p(r+e)2-pr2]/e

Try to take it from there.

5. Jan 29, 2004

### lastlaugh

another hint see what from: (pi(r+e)^2+pi*r^2)/h
you can factor out front and you will learn one of the fundamental problem solving techniques of derivatives.

6. Feb 1, 2004

### modmans2ndcoming

a derivative is....

all a derivative is, is the slope of a line that is tangent to a graph at a particular point P.

that might sound a bit complicated, but think of a circle.

if you draw a line that touches the circle at exactly one point, that line has an equasion. that equasion is the slope (change in y over the change in X) at that particular point. if you are at the very top or bottom of the circle, the derivative will be zero because the slope is zero on a horazontal line. if you are at the exact left or right of the circle, the derivative will be undefined because the slope of a vertical line is undefined.

as to the slope with respect to the radius, you would need to manipulate the equasion of the circle so that you are solving for X or Y so that you can take the derivative with respect to r. just a bit of algebra before you start is all.

7. Feb 1, 2004

### modmans2ndcoming

re:re:me again

I think "e" is not a very good variable to use when defining the derivative since "e" is a number and he/she will most certainly be using it in the next few weeks...no need to make it confusing for him/her.

I use "h", but that is just me.

8. Feb 1, 2004

### modmans2ndcoming

oh, also, as for easily solving

what meathods of derivation have you learned so far?

if you are just learning the formal definition, don't worry, it will get easier when you learn other ways to take a derivative....

if, you are beyond the formal definition, then it does not get much easier ;-)

9. Feb 2, 2004

### franz32

what meathods of derivation have you learned so far?

Well, I am already under the topic optimization... and the intermediate value theorem.. =)

10. Feb 12, 2004

### franz32

Thank you very much!

Hello guys, I really thanked all of you who help me about the derivatives. I learned it now. =)

11. Jul 23, 2004

### mathwonk

to take the derivative of an area of a circle wrt the radius, compare the change in area to the change in radius. Note that the change in area for a small change in radius is the area of a small collar or ring, which looks like a rectangle rolled up. the height of the rectangle is the change in radius and the length of the rolled up rectangle is the circumference, hence dividing by the change in radius gives the circumference. there fore the derivative of area wrt radius is circumference. to check this note that d/dr of pir^2 is 2pi r, which is the circumference.