We start with the derivative of a power function, $\ds f(x)=x^n$. Here $n$ is a number of any kind: integer, rational, positive, negative, even irrational, as in $\ds x^\pi$. 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 3.1.1 Find the derivative of $\ds f(x)=x^3$. $$\eqalign{ {d\over dx}x^3 &= \lim_{\Delta x\to0} {(x+\Delta x)^3-x^3\over \Delta x}.\cr &=\lim_{\Delta x\to0} {x^3+3x^2\Delta x+3x\Delta x^2 + \Delta x^3 -x^3\over \Delta x}.\cr &=\lim_{\Delta x\to0}{3x^2\Delta x+3x\Delta x^2 + \Delta x^3\over \Delta x}.\cr &=\lim_{\Delta x\to0}3x^2+3x\Delta x + \Delta x^2 = 3x^2.\cr }$$

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 $\ds (x+\Delta x)^n$ 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 $\ds x^i\Delta x^j$, and in fact that $i+j=n$ in every term. One way to see this is to understand that one method for multiplying out $\ds (x+\Delta x)^n$ 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 $\ds (x+\Delta x)^3$, 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 $\ds nx^{n-1}\Delta x$. 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 $\ds a_2$, $\ds a_3$, 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 3.1.2 Find the derivative of $\ds y=x^{-3}$. Using the formula, $\ds y'=-3x^{-3-1}=-3x^{-4}$.

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 $\ds f(x)=x^0$, though there is some question about just what this means at $x=0$. If we apply the power rule, we get $\ds f'(x)=0x^{-1}=0/x=0$, 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.

Exercises 3.1

Find the derivatives of the given functions.

Ex 3.1.1 $\ds x^{100}$ (answer)

Ex 3.1.2 $\ds x^{-100}$ (answer)

Ex 3.1.3 $\displaystyle {1\over x^5}$ (answer)

Ex 3.1.4 $\ds x^\pi$ (answer)

Ex 3.1.5 $\ds x^{3/4}$ (answer)

Ex 3.1.6 $\ds x^{-9/7}$ (answer)