Re: what "REALLY" is derivative?


Luke Wu wrote:
> Hi,
>
> Back in grade 11 when learning about parabolas, a teacher told me that
> there are lines that can touch the parabola only at one point, for
> every point on the curve (tangents), and that we needed to wait for
> more advanced courses to find the equation of these lines.
>
> I thought this was nonsense, because if two curves (parabola and line)
> intersect, then it's easy to solve for the slope of the line (because
> we knew how to solve the intersection of two lines- "this can't be much
> different" i thougth).. this is what I did:
>
> equation of parabola: y = ax^2 + bx + c
>
> point slope equation of line: y - y1 = m(x - x1)
>
> what I did was substitute the point (q , aq^2 + bq + c) for (x1,y1) in
> the slope equation
>
>
> y - aq^2 - bq - c = m(x - q)
>
> then substitute the y = aq^2 + bq + c
>
> ax^2 + bx + c - aq^2 - bq - c = m(x - q)
>
> a(x^2 - q^2) + b(x - q)= m(x - q)
>
> (x-q) [ a(x+q) + b ] = m(x - q)
>
> THEN I DID THE UNTHINKABLE and divided both sides by (x-q)
>
> m = a(x+q) + b and that is the slope of a line touching parabola at x
> = q
>
> I had no clue what a limit was, or a derivative, but I used the ideas
> learned about the intersection of two lines, and used it for the
> intersection of a parbola and a line (and just ASSUMED that x-q != 0)
>
> I then tried this for other curves successfully, and when I started
> learning calculus (2 years later), I couldn't get past the fact that
> the limit seemed to me (for reasons shown above) as an excuse to allow
> us to divide f(x) - f(a) by x - a.
>
> How is there always a x-a (for us to cancel out) within f(x) - f(a) if
> f is differentiable at 'a'. Does anyone know how this really works?
> Have any of you wondered how solving limits (be in with e/d or using
> limit laws/continuity without e/d) always works out perfect (we always
> get a factor x-a or h that cancels out or we always get the magical
> absolute value term with e-d limit proofs)? I know if f(x) - f(a) is a
> polynomial, it is zero when x = a, so there must be a factor x-a (by
> factor theorem) ... but really.. how "really" does this work?
>
> My post sounds unintelligible because I can't shape my
> confusions/thoughts properly to explain it clearly (and English is not
> my native tongue). My "funky" grade 11 self discovered method is
> clashing with the limit of difference quotient definition of a
> derivative. My extreme inquisitiveness (yes extreme, which is why I
> always do bad in school, because I can't get over the "WHYYYYY?" even
> if I"m told the WHYYY?? is coming later on).
>
> Can anyone shed some light on my confusion? What "really" is a
> derivative?
>
> Any insights or guidance would be greatly appreciated.

In a sense, your are correct when you say that taking a limit is an
excuse to divide f(x) - f(a) by x - a. In fact, limits allow us to do
many things we couldn't do with the basic algebraic operations. A
simple but typical example of this is that we can't divide by zero, but
we may still want to see what happens when we get arbitrarily close to
zero. Examples of this are 1/(x^2) and (sin x)/x.

While it is nice that you were able to find a solution without using
limits, your method is very limited. You might have had some success
with polynomials, but have you tried finding derivatives of functions
like sin x or e^x without using calculus/analysis of some kind? Along
the same lines, these are very good examples of functions where a
factor of (x - a) doesn't just pop out when computing the derivative.

There is a fairly simple geometric argument for what is going on when
computing the derivative that you may or may not know. Just draw a
curve on a piece of paper and call your curve f(x). For any real
number a, you have a point (a, f(a)) on your curve and you can call any
point on your curve (x, f(x)). Now, what would the slope between these
two points be? Using simple formulas you learned in your 11th grade
class or earlier, the slope is (f(x) - f(a))/(x - a). So the line
connecting these two points has this slope, and these lines are
sometimes called secant lines in calculus books. Initially, this line
will intersect your curve in both of the points (a, f(a)) and (x,
f(x)). This is where the limit comes in. As you let x approach a,
these points become arbitrarily close to one another so that you get
the slope of the tangent line at x = a. So the limit definition of the
derivative is actually pretty clever in that it uses the simple fact
that we can find the slope of a line connecting any two points to find
the slope of a line through a single point tangent to the curve. Of
course, this is only possible because of the limiting procedure used.

> THEN I DID THE UNTHINKABLE and divided both sides by (x-q)

As a quick comment about this, what you did was technically an illegal
operation and completely unnecessary in your calculation. Here's what
you can do instead:

a(x^2 - q^2) + b(x - q) = m(x - q)

a(x^2 - q^2) + b(x - q) - m(x - q) = 0

(x - q) [a(x + q) + b - m] = 0.

Solving for zero in the second quantity gives m = a(x + q) + b, exactly
as you got before.

Anyway, I hope this might make a little more sense. If not, there's
plenty of other people on here who can give it a shot.

Mike

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