## Monday, June 24, 2013

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

Posted by: Rebekah

Okay, we have seen that the derivative can be defined as follows: $\underset{dx\to 0}{lim}\frac{f\left(x+dx\right)-f\left(x\right)}{dx}$

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 $cn{x}^{n-1}$.

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:  $\underset{dx\to 0}{lim}\frac{c\left(x+dx{\right)}^{n}-c{x}^{n}}{dx}$

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!
$\underset{dx\to 0}{lim}\frac{c\left(x+dx{\right)}^{n}-c{x}^{n}}{dx}=$
$\underset{dx\to 0}{lim}c\left[{x}^{n}+n{x}^{n-1}\cdot dx+\frac{n!}{2!\left(n-2\right)!}{x}^{n-2}\cdot d{x}^{2}+...+nx\cdot d{x}^{n-1}+d{x}^{n}\right]\cdot \frac{1}{dx}-\frac{c{x}^{n}}{dx}$

Now let's multiply through by the c located outside the brackets:
$\underset{dx\to 0}{lim}\left[c{x}^{n}+cn{x}^{n-1}\cdot dx+c\frac{n!}{2!\left(n-2\right)!}{x}^{n-2}\cdot d{x}^{2}+...+cnx\cdot d{x}^{n-1}+cd{x}^{n}-c{x}^{n}\right]\frac{1}{dx}$

Notice that I was able to place the -cxⁿ inside the bracket because it too was being multiplied by  $\frac{1}{dx}$ . Consequently, the cxⁿ in the front cancels with the -cxⁿ at the end.
$\underset{dx\to 0}{lim}\left[cn{x}^{n-1}\cdot dx+c\frac{n!}{2!\left(n-2\right)!}{x}^{n-2}\cdot d{x}^{2}+...+cnx\cdot d{x}^{n-1}+cd{x}^{n}\right]\frac{1}{dx}$

Hey look! Each term inside the brackets has a "dx", so we can factor the "dx" out!
$\underset{dx\to 0}{lim}\left[cn{x}^{n-1}+c\frac{n!}{2!\left(n-2\right)!}{x}^{n-2}\cdot dx+...+cnx\cdot d{x}^{n-2}+cd{x}^{n-1}\right]\frac{dx}{dx}$

Which is the same as...
$\underset{dx\to 0}{lim}\left[cn{x}^{n-1}+c\frac{n!}{2!\left(n-2\right)!}{x}^{n-2}\cdot dx+...+cnx\cdot d{x}^{n-2}+cd{x}^{n-1}\right]$

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

Thus, we find that the derivative of any function of the form cxⁿ is indeed $cn{x}^{n-1}$

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