Derivative

[stopandthink]

[math] f(x)= x^2[/math]

[math] f(4)=16[/math]

[math] f'(4)=8[/math] What exactly is the derivative giving me?

[timo]

I think that a question as broad and non-specialized as "what is a derivative" is better suited for Google and Wikipedia than for a discussion forum. The derivative is the local trend of a graph, but there's arbitrarily more to say about it. So I strongly advice reading the Wikipedia article or your school book (if you have one).

Edited November 1, 2012 by timo

[stopandthink]

I know that it's when a secant line gets closer and closer to the tangent line at the co-ordinates (4, 16) ... so when we get really close we get 8 as a "derivative"... but i still know nothing about a derivative.

Edited November 1, 2012 by stopandthink

[timo]

You could know that it is the local trend of the function - I just told you so. But let's try it differently: What kind of answer do you expect?

[JohnStu]

[math] f(x)= x^2[/math]

[math] f(4)=16[/math]

[math] f'(4)=8[/math] What exactly is the derivative giving me?

 

 

that derivative is giving the instantaneous rate of change at the point when x=4 of the function f(x), which is 8, as shown

[stopandthink]

I'm not expecting a different answer because i know how to find the "derivative", Of this simple function.

 

Ok, so i think of a derivative as rise/run=slope... and i can find the slope of a secant line easily but what we want to do is get

the secant line as close to the tangent line.

So when we do arrive at the tangent line, we find that it's 8.... but i have no clue as to what i'm looking at on a graph with this number..

[ydoaPs]

I'm not expecting a different answer because i know how to find the "derivative", Of this simple function.

 

Ok, so i think of a derivative as rise/run=slope... and i can find the slope of a secant line easily but what we want to do is get

the secant line as close to the tangent line.

So when we do arrive at the tangent line, we find that it's 8.... but i have no clue as to what i'm looking at on a graph with this number..

Your tangent line has the slope that is the derivative. Now, you've already got an x value, a y value, and the derivative of the function at that x value. Now all you need to do is take that information and find the equation of a line that satisfies it. It may be easiest to start with point-slope form and solve for y.

[timo]

A slope of 8 means that at this point if you increase the x value by some small about A, then the y-value will change by roughly 8*A. The smaller A is, the better this approximation holds true (technically it is defined as being exactly 8*A for arbitrarily small A, but that's too technical to be helpful).

 

Somewhat more abstract - and possibly making the whole background more clear: A function gives you a relation between two properties. For a given pair of values for these two properties, the derivative gives you information how one of the properties behaves if the other one is slightly changed.

 

There is one technicality that possibly make derivatives seem less useful to you than they actually are: The whole issue with derivatives becomes really interesting if you don't know the actual function (such cases are very common in physics, for example) or if the function is too complicated to handle. Of course, those are not the cases on which you introduce derivatives in school. So I do have some sympathy for not immediately seeing the point in derivatives.

[randomc]

I think of it as where the value for the slope of the function on the graph is going and also where it's been, i.e, the local trend of the slope (per Timo). The value '8' in your example is an average of where it's been and where it's going.

 

I think it can be useful for predicting or recovering data you can't (or don't feel like) directly looking at.

[stopandthink]

Ok so i just learned that f'(4)=8

is the point on a new graph f'(x) that overlaps the original graph f(x), which is why i couldn't understand where it belonged...

[ydoaPs]

Ok so i just learned that f'(4)=8

is the point on a new graph f'(x) that overlaps the original graph f(x)

 

I'm having trouble finding an interpretation of this that is correct.

[stopandthink]

I'm having trouble finding an interpretation of this that is correct.

Edited November 2, 2012 by stopandthink

[mississippichem]

[math] f(x)= x^2[/math]

[math] f(4)=16[/math]

[math] f'(4)=8[/math] What exactly is the derivative giving me?

 

I think timo already explained this pretty well, but I'll attempt a rephrase in the hope that it helps you in some way.

 

You can just think of the derivative loosely as a generalization of slope to continuous functions. The concept of slope really only applies to straight lines, y=mx+b where "m" is the slope. The derivative just takes that concept and applies it to a larger class of functions.

 

To really understand it you'll need a pretty good sense of what a limit is. Without limits all you can really do is hand wave. Do you understand that the derivative is really just a limit of the difference quotient as the change in the function goes to zero?

 

Some people benefit from a more rigorous treatment of mathematics. It seems counter-intuitive but sometimes going a little deeper can pay off great returns in terms of more mature understanding.

[stopandthink]

Ok, i think i understand better now what the derivative is. By drawing out a graph [math]f(x)=x^2[/math] with x=time(in seconds), y=velocity(mph)

 

So when [math] \frac{2seconds} {4mph}[/math], [math] \frac{3seconds} {9mph}[/math] So the difference is 5mph, but as you get closer and closer to exactly 3 seconds you find that it's instantaneous velocity(derivative) is 6mph...?