Power rule from limit definition (positive integer exponents)

Calculus Tutorial

Power rule from limit definition (positive integer exponents)

Intro

We will attempt to derive the power rule for derivatives from the limit definition of the derivative. For the purposes of this problem we will limit ourselves to positive integer exponents.

Sample Problem

Find the derivative of x^n, for positive integers n, using the limit definition of a derivative. Hint: It may be useful to recall the binomial theorem.

Solution

We begin by writing the limit definition of a derivative:

    \[ f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}. \]

In our case f(x)=x^n so this becomes

    \[ (x^n)'=\lim_{h\rightarrow 0} \frac{(x+h)^n-x^n}{h}. \]

In order to solve this limit is will be necessary to expand the (x+h)^n term. To do this, we will use the binomial theorem.

The binomial theorem is given by the following equation:

    \[ (x+h)^n=x^n+\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2+\ldots+\binom{n}{n-1}x^{1}h^{n-1}+h^n, \]

where

    \[ \binom{n}{k}=\frac{n!}{k!(n-k)!}. \]

We do not need to know these coefficients yet, so for now represent them as c_1,c_2,\ldots,c_{n-1}. Thus we have

    \[ (x+h)^n=x^n+c_1x^{n-1}h+c_2x^{n-2}h^2+\ldots+c_{n-1}x^{1}h^{n-1}+h^n. \]

Finally, we plug this into the limit and solve. Plugging this equation into the limit, we get

    \[ \lim_{h\rightarrow 0} \frac{(x+h)^n-x^n}{h} = \lim_{h\rightarrow 0} \frac{x^n+c_1x^{n-1}h+c_2x^{n-2}h^2+\ldots+c_{n-1}x^{1}h^{n-1}+h^n-x^n}{h}. \]

We can now cancel out the two x^n terms in the numerator and then divide out an h from each of the remaining terms. We are left with

    \[ (x^n)'=\lim_{h\rightarrow 0} c_1x^{n-1}+c_2x^{n-2}h+\ldots+c_{n-1}x^{1}h^{n-2}+h^{n-1}. \]

We can now substitute in h=0, and note that only the first term remains. So

    \[ (x^n)'=c_1x^{n-1}. \]

Recall from above that c_1=\binom{n}{1}, but we can reduce this as follows:

    \[ \begin{aligned} c_1&=\binom{n}{1}\\ &=\frac{n!}{1!(n-1)!}\\ &=\frac{n(n-1)(n-2)\cdots(2)(1)}{1(n-1)(n-2)\cdots(2)(1)}\\ &=n. \end{aligned} \]

Hence, our final answer is

    \[ (x^n)'=nx^{n-1}, \]

which is exactly the power rule.



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