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)