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

Hey look! Each term inside the brackets has a "dx", so we can factor the "dx" out!
$$\underset{dx\to 0}{lim}[cn{x}^{n-1}+c\frac{n!}{2!(n-2)!}{x}^{n-2}\cdot dx+...+cnx\cdot d{x}^{n-2}+cd{x}^{n-1}]\frac{dx}{dx}$$
Which is the same as...
$$\underset{dx\to 0}{lim}[cn{x}^{n-1}+c\frac{n!}{2!(n-2)!}{x}^{n-2}\cdot dx+...+cnx\cdot d{x}^{n-2}+cd{x}^{n-1}]$$$$$$
It becomes clear that, as x becomes exceedingly small (so much so that it approximates zero), all other terms, with the exception of $cn{x}^{n-1}$, 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!!