Silly differentiation from first principles question

[Monarch]

I've got some maths homework to do over the summer before I go back to uni and there's this stupid question on there which is one of those 'so basic I don't know it' kind of questions, so here goes.

Homework Statement

What is Lim_Δx→0[itex]\frac{y(x+Δx) - y(x)}{Δx}[/itex] ?

The Attempt at a Solution

I know that this is just differentiation from first principles, you can take that and derive the whole d/dx relation and stuff, but what exactly is this?

 

[uart]

Yes it's differentiation from first principles. You can think of it as the limit of [itex]\frac{{\rm rise}}{{\rm run}}[/itex] as "run" goes to zero. It therefore is a generalization of the concept of the gradient of a straight line to the "incremental gradient" of a curve.

Why don't you have an attempt at evaluating the limit for some simple functions, like [itex]y=3x+2[/itex], [itex]y=x^2[/itex] and [itex]y=1/x[/itex] for some simple examples.

 

[voko]

If that is all of the question, then it is really a simple question. What is the technical term for this particular kind of limit? Differentiation is not exactly right, it explains what you are doing, but does not name the result.

 

[LCKurtz]

I've got some maths homework to do over the summer before I go back to uni and there's this stupid question on there which is one of those 'so basic I don't know it' kind of questions, so here goes.

Homework Statement

What is Lim_Δx→0[itex]\frac{y(x+Δx) - y(x)}{Δx}[/itex] ?

The Attempt at a Solution

I know that this is just differentiation from first principles, you can take that and derive the whole d/dx relation and stuff, but what exactly is this?

The quantity you are taking the limit of is called the difference quotient. The result of taking the limit of the difference quotient is the derivative y'(x).

 

[Monarch]

Thanks all, like I said, so basic I don't really know what to say about it. I use that kind of thing so often but never have to think about exactly what it is that is seems I've just taken its existence for granted, but I should be able to get an answer out of it from the comments made.