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# 3.1: The Power Rule

• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College

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We start with the derivative of a power function, $$f(x)=x^n$$. Here $$n$$ is a number of any kind: integer, rational, positive, negative, even irrational, as in $$x^\pi$$. We have already computed some simple examples, so the formula should not be a complete surprise:

${d\over dx}x^n = nx^{n-1}.$

It is not easy to show this is true for any $$n$$. We will do some of the easier cases now, and discuss the rest later. The easiest, and most common, is the case that $$n$$ is a positive integer. To compute the derivative we need to compute the following limit:

${d\over dx}x^n = \lim_{\Delta x\to0} {(x+\Delta x)^n-x^n\over \Delta x}.$

For a specific, fairly small value of $$n$$, we could do this by straightforward algebra.

Example $$\PageIndex{1}$$

Find the derivative of $$f(x)=x^3$$.

Solution

\eqalign{ {d\over dx}x^3 &= \lim_{\Delta x\to0} {(x+\Delta x)^3-x^3\over \Delta x}.\cr& =\lim_{\Delta x\to0} {x^3+3x^2\Delta x+3x\Delta x^2 + \Delta x^3 -x^3\over \Delta x}.\cr& =\lim_{\Delta x\to0}{3x^2\Delta x+3x\Delta x^2 + \Delta x^3\over \Delta x}.\cr& =\lim_{\Delta x\to0}3x^2+3x\Delta x + \Delta x^2 = 3x^2.\cr }

The general case is really not much harder as long as we don't try to do too much. The key is understanding what happens when $$(x+\Delta x)^n$$ is multiplied out:

$(x+\Delta x)^n=x^n + nx^{n-1}\Delta x + a_2x^{n-2}\Delta x^2+\cdots+ +a_{n-1}x\Delta x^{n-1} + \Delta x^n.$

We know that multiplying out will give a large number of terms all of the form $$x^i\Delta x^j$$, and in fact that $$i+j=n$$ in every term. One way to see this is to understand that one method for multiplying out $$(x+\Delta x)^n$$ is the following: In every $$(x+\Delta x)$$ factor, pick either the $$x$$ or the $$\Delta x$$, then multiply the $$n$$ choices together; do this in all possible ways. For example, for $$(x+\Delta x)^3$$, there are eight possible ways to do this:

\eqalign{ (x+\Delta x)(x+\Delta x)(x+\Delta x)&=xxx + xx\Delta x + x\Delta x x + x\Delta x \Delta x\cr& \qquad+ \Delta x xx + \Delta xx\Delta x + \Delta x\Delta x x + \Delta x\Delta x \Delta x\cr& = x^3 + x^2\Delta x +x^2\Delta x +x\Delta x^2\cr& \quad+x^2\Delta x +x\Delta x^2 +x\Delta x^2 +\Delta x^3\cr& =x^3 + 3x^2\Delta x + 3x\Delta x^2+\Delta x^3\cr }

No matter what $$n$$ is, there are $$n$$ ways to pick $$\Delta x$$ in one factor and $$x$$ in the remaining $$n-1$$ factors; this means one term is $$nx^{n-1}\Delta x$$. The other coefficients are somewhat harder to understand, but we don't really need them, so in the formula above they have simply been called $$a_2$$, $$a_3$$, and so on. We know that every one of these terms contains $$\Delta x$$ to at least the power 2.

Now let's look at the limit:

\eqalign{ {d\over dx}x^n &= \lim_{\Delta x\to0} {(x+\Delta x)^n-x^n\over \Delta x}\cr& =\lim_{\Delta x\to0} {x^n + nx^{n-1}\Delta x + a_2x^{n-2}\Delta x^2+\cdots+ a_{n-1}x\Delta x^{n-1} + \Delta x^n-x^n\over \Delta x}\cr& =\lim_{\Delta x\to0} {nx^{n-1}\Delta x + a_2x^{n-2}\Delta x^2+\cdots+ a_{n-1}x\Delta x^{n-1} + \Delta x^n\over \Delta x}\cr& =\lim_{\Delta x\to0} nx^{n-1} + a_2x^{n-2}\Delta x+\cdots+ a_{n-1}x\Delta x^{n-2} + \Delta x^{n-1} = nx^{n-1}.\cr }

Now without much trouble we can verify the formula for negative integers. First let's look at an example:

Example $$\PageIndex{2}$$

Find the derivative of $$y=x^{-3}$$. Using the formula, $$y'=-3x^{-3-1}=-3x^{-4}$$.

Solution

Here is the general computation. Suppose $$n$$ is a negative integer; the algebra is easier to follow if we use $$n=-m$$ in the computation, where $$m$$ is a positive integer.

\eqalign{ {d\over dx}x^n &= {d\over dx}x^{-m} = \lim_{\Delta x\to0} {(x+\Delta x)^{-m}-x^{-m}\over \Delta x}\cr& =\lim_{\Delta x\to0} { {1\over (x+\Delta x)^m} - {1\over x^m} \over \Delta x} \cr& =\lim_{\Delta x\to0} { x^m - (x+\Delta x)^m \over (x+\Delta x)^m x^m \Delta x} \cr& =\lim_{\Delta x\to0} { x^m - (x^m + mx^{m-1}\Delta x + a_2x^{m-2}\Delta x^2+\cdots+ a_{m-1}x\Delta x^{m-1} + \Delta x^m)\over (x+\Delta x)^m x^m \Delta x} \cr& =\lim_{\Delta x\to0} { -mx^{m-1} - a_2x^{m-2}\Delta x-\cdots- a_{m-1}x\Delta x^{m-2} - \Delta x^{m-1})\over (x+\Delta x)^m x^m} \cr& ={ -mx^{m-1} \over x^mx^m}= { -mx^{m-1} \over x^{2m}}= -mx^{m-1-2m}= nx^{-m-1} = nx^{n-1}.\cr }

We will later see why the other cases of the power rule work, but from now on we will use the power rule whenever $$n$$ is any real number. Let's note here a simple case in which the power rule applies, or almost applies, but is not really needed. Suppose that $$f(x)=1$$; remember that this "1'' is a function, not "merely'' a number, and that $$f(x)=1$$ has a graph that is a horizontal line, with slope zero everywhere. So we know that $$f'(x)=0$$. We might also write $$f(x)=x^0$$, though there is some question about just what this means at $$x=0$$. If we apply the power rule, we get $$f'(x)=0x^{-1}=0/x=0$$, again noting that there is a problem at $$x=0$$. So the power rule "works'' in this case, but it's really best to just remember that the derivative of any constant function is zero.

## Contributors

• Integrated by Justin Marshall.