Monday, June 24, 2013

Math Monday: Derivatives of Exponential Functions--the Power Rule

 Posted by: Rebekah

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Okay, we have seen that the derivative can be defined as follows: lim d x 0 f ( x + d x ) f ( x ) d x

Please note that the only difference here is that I am using "dx" as a small increase rather than "da". 

Also, if we attempt to find a pattern for such derivatives (e.g x², x³), we find the following (please see Table 1.1):

                                                                                                                   Table 1.1

In other words, it appears that, if a function is equal to c x n, then its derivative must be equal to c n x n 1.

But how can we prove this to be true?

Well, using the definition of a derivative above, we are lead to conclude that the derivative must be equal to:  lim d x 0 c ( x + d x ) n c x n d x

Now all there is left to do is solve!

Before solving, please note the expression  c( x + dx )ⁿ : there is no doubt this will require the binomial theorem.

                                                                      Image from: mathisfun.com

Okay, now we can start solving!
lim d x 0 c ( x + d x ) n c x n d x =
lim d x 0 c [ x n + n x n 1 d x + n ! 2 ! ( n 2 ) ! x n 2 d x 2 + . . . + n x d x n 1 + d x n ] 1 d x c x n d x

Now let's multiply through by the c located outside the brackets:
lim d x 0 [ c x n + c n x n 1 d x + c n ! 2 ! ( n 2 ) ! x n 2 d x 2 + . . . + c n x d x n 1 + c d x n c x n ] 1 d x

Notice that I was able to place the -cxⁿ inside the bracket because it too was being multiplied by  1 d x . Consequently, the cxⁿ in the front cancels with the -cxⁿ at the end.
lim d x 0 [ c n x n 1 d x + c n ! 2 ! ( n 2 ) ! x n 2 d x 2 + . . . + c n x d x n 1 + c d x n ] 1 d x

Hey look! Each term inside the brackets has a "dx", so we can factor the "dx" out!
lim d x 0 [ c n x n 1 + c n ! 2 ! ( n 2 ) ! x n 2 d x + . . . + c n x d x n 2 + c d x n 1 ] d x d x

Which is the same as...
lim d x 0 [ c n x n 1 + c n ! 2 ! ( n 2 ) ! x n 2 d x + . . . + c n x d x n 2 + c d x n 1 ]

It becomes clear that, as x becomes exceedingly small (so much so that it approximates zero), all other terms, with the exception of    c n x n 1 , also approximate zero.

Thus, we find that the derivative of any function of the form cxⁿ is indeed  c n x n 1 

Woohoo!!! This means our assumptions were correct!!