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Proof of Derivative for Function f(x) = ax^n

Date: 01/14/2004 at 20:44:47
From: Colin
Subject: proof of the "quick form" of derivitives of A*X^N

I've been wondering if there is a proof for the "quick form" of the
derivative in the ax^n case?  We just learned it after using limits to
calculate the derivatives.  I like the quick method, but I'm the kind
of person who likes to know why and how things work.

Date: 01/15/2004 at 09:09:15
From: Doctor Peterson
Subject: Re: proof of the "quick form" of derivitives of A*X^N

Hi, Colin.

We first define derivatives using limits, then we apply that
definition to find simple rules for the derivatives of common
functions, and then rarely go back to the definition again.  The main
value of the definition is to allow us to prove the rules and other
theorems about derivatives.

Let's look at the function f(x) = ax^n. The derivative is

       f(x+h) - f(x)        a(x+h)^n - ax^n
  lim  ------------- = lim  ---------------
  h->0       h         h->0        h

At this point you need the binomial expansion; among other places, you
can find this discussed in our FAQ on Pascal's triangle.  Or see

  Binomial Expansions and Pascal's Triangle 

All that matters to us is the first two terms:

  (a+b)^n = a^n + n a^(n-1) b + n(n-1)/2 a^(n-2) b^2 + ...

where the rest of the terms have whole coefficients with decreasing 
powers of a and increasing powers of b.  Setting a = x and b = h, and 
putting this into the derivative, we get

       a[x^n + nx^(n-1)h + n(n-1)/2 x^(n-2) h^2 + ...] - ax^n
  lim  ------------------------------------------------------
  h->0                          h

Note that the first term of the expansion will cancel with the -ax^n
at the end, leaving

       anx^(n-1)h + an(n-1)/2 x^(n-2) h^2 + ...
  lim  ----------------------------------------
  h->0                   h

Up to this point we still have the form 0/0, which means there's more 
to do.  But now we can divide by h.  The unshown terms above all have
a factor of at least h^2, so we get

   = lim [anx^(n-1) + h(an(n-1)/2 x^n-2 + ...) ]

   = anx^(n-1)

since the term with a factor of h goes to zero.

The main idea of limits is that when we simplify a function, as by 
dividing by h here, we get a continuous function that is equivalent to
the original everywhere except where the latter was not defined;
therefore the new function's VALUE at that point is the same as the 
LIMIT of the original function.  In effect, we are "filling in the 
hole".  Not all limits can be solved that easily, but when it can be 
done, it makes the work very easy.  And now that we've done it, we 
don't need to bother with limits when we need to find the derivative 
of a polynomial.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 

Date: 01/20/2004 at 15:54:01
From: Colin
Subject: Thank you (proof of the "quick form" of derivitives of A*X^N)

Thank you very much!  This is very cool, and now I understand why the
'quick method' actually works.  Thanks for taking the time to answer
my question.
Associated Topics:
High School Calculus

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