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3
Rules for Finding Derivatives




It is tedious to compute a limit every time we need to know the derivative of a function.
Fortunately, we can develop a small collection of examples and rules that allow us to
compute the derivative of almost any function we are likely to encounter. Many functions
involve quantities raised to a constant power, such as polynomials and more complicated
combinations like y = (sin x)4 . So we start by examining powers of a single variable; this
gives us a building block for more complicated examples.


3.1 The Power Rule

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

d n
x = nxn−1 .
dx

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 n (x + ∆x)n − xn
x = lim .
dx ∆x→0 ∆x

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

55

,56 Chapter 3 Rules for Finding Derivatives

EXAMPLE 3.1.1 Find the derivative of f (x) = x3 .

d 3 (x + ∆x)3 − x3
x = lim .
dx ∆x→0 ∆x
x3 + 3x2 ∆x + 3x∆x2 + ∆x3 − x3
= lim .
∆x→0 ∆x
3x2 ∆x + 3x∆x2 + ∆x3
= lim .
∆x→0 ∆x
= lim 3x2 + 3x∆x + ∆x2 = 3x2 .
∆x→0


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 + ∆x)n is multiplied out:

(x + ∆x)n = xn + nxn−1 ∆x + a2 xn−2 ∆x2 + · · · + +an−1 x∆xn−1 + ∆xn .

We know that multiplying out will give a large number of terms all of the form xi ∆xj , 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 + ∆x)n is the following: In every (x + ∆x) factor, pick either the x
or the ∆x, then multiply the n choices together; do this in all possible ways. For example,
for (x + ∆x)3 , there are eight possible ways to do this:

(x + ∆x)(x + ∆x)(x + ∆x) = xxx + xx∆x + x∆xx + x∆x∆x
+ ∆xxx + ∆xx∆x + ∆x∆xx + ∆x∆x∆x
= x3 + x2 ∆x + x2 ∆x + x∆x2
+ x2 ∆x + x∆x2 + x∆x2 + ∆x3
= x3 + 3x2 ∆x + 3x∆x2 + ∆x3
No matter what n is, there are n ways to pick ∆x in one factor and x in the remaining
n−1 factors; this means one term is nxn−1 ∆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 a2 , a3 , and so on. We know that every one of these terms contains ∆x to at
least the power 2. Now let’s look at the limit:
d n (x + ∆x)n − xn
x = lim
dx ∆x→0 ∆x
x + nxn−1 ∆x + a2 xn−2 ∆x2 + · · · + an−1 x∆xn−1 + ∆xn − xn
n
= lim
∆x→0 ∆x
n−1 n−2
nx ∆x + a2 x ∆x + · · · + an−1 x∆xn−1 + ∆xn
2
= lim
∆x→0 ∆x
= lim nxn−1 + a2 xn−2 ∆x + · · · + an−1 x∆xn−2 + ∆xn−1 = nxn−1 .
∆x→0

, 3.1 The Power Rule 57

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 y = x−3 . Using the formula, 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.

d n d −m (x + ∆x)−m − x−m
x = x = lim
dx dx ∆x→0 ∆x
1 1
(x+∆x)m − xm
= lim
∆x→0 ∆x
x − (x + ∆x)m
m
= lim
∆x→0 (x + ∆x)m xm ∆x

xm − (xm + mxm−1 ∆x + a2 xm−2 ∆x2 + · · · + am−1 x∆xm−1 + ∆xm )
= lim
∆x→0 (x + ∆x)m xm ∆x
−mxm−1 − a2 xm−2 ∆x − · · · − am−1 x∆xm−2 − ∆xm−1 )
= lim
∆x→0 (x + ∆x)m xm
−mxm−1 −mxm−1
= = = −mxm−1−2m = nx−m−1 = nxn−1 .
xm xm x2m

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) = x0 , 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.

Exercises 3.1.
Find the derivatives of the given functions.
1. x100 ⇒ 2. x−100 ⇒
1
3. ⇒ 4. xπ ⇒
x5
5. x3/4 ⇒ 6. x−9/7 ⇒

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