# Generic definition of derivative?

1. Apr 12, 2012

### Hernaner28

Generic "definition" of derivative?

Hi. This is a theoric doubt I have since I went to class today. The professor "redifined" the derivative at point a. He draw a curve (the function) and the tangent at point a. Then he draw another two lines in the same point.
Well, then he said that the error is the difference between the lines and the function or I understood that (if it's not that way please clarify that to me).

So:

\eqalign{ & Error = f(x) - y \cr & Error = f(x) - [m(x - a) + f(a)] \cr & Error = f(x) - f(a) - m(x - a) \cr & Error = \underbrace {(x - a)}_{\scriptstyle 0 \atop \scriptstyle x \to a} \underbrace {\underbrace {\left( {\frac{{f(x) - f(a)}}{{x - a}} - m} \right)}_{\scriptstyle f'(a) - m \atop \scriptstyle x \to a} }_{r(x)} \cr & {\text{The error is minimum if m = f'(a)}} \cr & {\text{so}} \cr & \underbrace {f(x) = f(a) + f'(a)(x - a) + (x - a)r(x)}_{{\text{With }}r(x) \to 0{\text{ when }}x \to a} \cr}

Now... this is really intresting but I wonder what this is for... is this useful for sth in particular?

Thanks!

2. Apr 12, 2012

### Hurkyl

Staff Emeritus
Re: Generic "definition" of derivative?

He's shown you two things.

First, he's shown you an interesting formula involving derivatives. This actually has quite a lot of practical use for coming up with good approximations, both in numeric and symbolic contexts. e.g. things like $\sin x \approx x$ for small x, or to quickly estimate the value of $\sqrt{101}$.

It may not seem like a big deal now. One place where it really becomes useful when you are dealing with more complicated functions and expressions; knowing this formula and when to use it can allow you to replace the complicated function with something much simpler and easier to analyze.

Also, it turns out that this definition of the derivative generalizes well to other contexts, such as when you start doing multi-variable calculus.

The second thing he's shown you is a useful technique for computing / defining things. It's a surprisingly tricky issue to give a good definition of the term "tangent line" or to compute such a thing. But geometrically, we see that the tangent line can be approximated by a secant line. Turning that around, secant lines are approximated by the tangent line, and we can use this property to define the tangent line, as the one that gives the best approximation to the secant lines.

This technique is one of the main techniques of differential calculus: replace the complicated idea with a simple idea that approximates it. Show the approximations are well-behaved. Then, define the complicated idea to be the thing that the simple idea is approximating.

3. Apr 12, 2012

### Hernaner28

Re: Generic "definition" of derivative?

Oh yes, it seems really intresting. So, this could be really called another definition if we say?:

$$f'(a) = \mathop {\lim }\limits_{\scriptstyle x \to a \atop \scriptstyle r(x) \to 0} \frac{{f(x) - f(a) - (x - a)r(x)}}{{x - a}}$$

I thought this was not a correct definition but now I see it's the same as:

$$f'(a) = \mathop {\lim }\limits_{x \to a} \frac{{f(x) - f(a)}}{{x - a}}$$

But I still haven't realise when it's convenient to use it. I guess I will learn it eventually.
Thank you!

4. Apr 12, 2012

### Hernaner28

Re: Generic "definition" of derivative?

Does this define the tangent or the derivative at a? I feel this is gonna be really useful to know for my course so I want to understand it now before it becomes dense.

5. Apr 13, 2012

### Hernaner28

Re: Generic "definition" of derivative?

I got confused again. I don't know what this defines, if the derivative or the tangent and I don't see any differences between this and the formal definition.

Wikipedia shows this:
$$\lim_{h\to 0}\frac{f(a+h)-f(a) - f'(a)\cdot h}{h} = 0$$

But I cannot connect this expresion to the other one! Here there's no r(x)

6. Apr 13, 2012

### Hurkyl

Staff Emeritus
Re: Generic "definition" of derivative?

It works like this:

Let $r_m(x)$ be the function
$$r_{m,a}(x) = \frac{ f(x) - f(a) - m (x-a) }{x-a}$$
Then we say $f'(a) = m$ if and only if
$$\lim_{x \to a} r_{m,a}(x) = 0$$​

It can be rearranged a fair bit. The above is the same as saying

We say $f'(a) = m$ if and only of
$$\lim_{x \to a} \frac{ f(x) - f(a) - m (x-a) }{x-a} = 0$$​