Example 2.5 To prove that f™(3) = 6, we need to show that
(3 + hJ2 ˜ 32 _ 6 as h, _ (,
(6)
hn
for every sequence {h,} which approaches zero, not just for the sequence (5).
We now prove (6) analytically. For any h,
(3 + h)* 3™ = 9+6h+h2p9 46 + h) = 6 + h
h
h h
which clearly converges to 6 as h  0. Now, we know for sue that f™(3) = 6.
Example 2.6 Now, add one more degree of generality and compute the derivative
1of f(n) = x2 at an arbitrary point x0. Let {h,} be an arbitrary sequence which
converges to 0 as n  m. Then,
f(,q, + h,,)  /.(I”) = (X” + h$  x; _ x; + Zh,,X” + /I,:  .$

hn hn hn
= k&h + U = 2wo + h
n.
hn
which tends to 2.x,,ash,,  0. This calculation proves the following theorem.
f(x) = x2 at x0 is f™(q) =
Theorem 2.2 The derivative of
Theorem 2.2 and Exercise 2.10 can be summarized by the statement that the
derivative of x1 is kxk ™ fork = 0, I, 2, 3, 4. WC next prove that this statement is
true for all positive integers k. Later, we™ll see that it is true for every real number
k, including negative numbers and fractions. In the proof of Theorem 2.2 and in
the proofs in part h of Exercise 2. IO, we used the explicit formula for (x + h)k for
small integers k. To prove the more general result. we need the general formula for
(x + hp for any positive integer k, a formula WE present in the following lemma.
Its proof can be found in any precollege algebra text under “binomial expansion.“
For any positive integer k.
Lemma 2 . 1
( x + h)k = xk + a,.,? ˜h™ + + axm,x™hk˜m™ + a&, (7)
k!
where f o r j = l.....k
” = j! (k ,j)!™
In particular, a, = k. a2 = k(k  1)/Z, and uk = 1
12.41 COMPUTING DERIVATIVES 27
For any positive integer k, the derivative of f(x) = xk at x0 is
Theorem 2.3
f™(x,,) = kx˜.
I
= h(k.$ + ;k(k  l)$˜h + + a&˜)
h
= k.$™ + ;k(k  l)$˜h + + a&˜,
which approaches Ax;™ as h  0. n
Rules for Computing Derivatives
The monomials X˜ are the basic building blocks for a large class of functions,
including all polynomials and rational functions. To compute the derivatives of
functions in these larger classes, we need to know how to take the derivative of a
sum; difference, product, or quotient of two functions whose derivatives we know
how to compute. First, recall that we add, subtract, divide. and multiply functions
in the natural way just by performing these operations on the values of the
functions. For example. if f (x) = x3 and &) = 6x™. then the sum, product, and
quotiem functions constructed from these two are, respectively:
(f + g)(x) = f(x) + R(X) = x3 + Ox˜,
(f g)(x)  f(x) g(x) = .x3 6x™ = 6x™.
The following theorem presents the rules for differentiating the sum, differ
ence. product, quotient. and power of functions. These rules, along with Theorem
2.3. will allow us to compute the derivatives of most clcmentary functions, includ
ing all polynomials and rational functions.
Part c of Theorem 2.4 is called the Product Rule, part rf the Quotient Rule,
and part e the Power Rule. Note that the derivative behaves very nicely with
respect to sums and dificrcnces of functions, but the rules for differentiating
products and quotients are a bit more complicated. The proof of each statement in
Thcorcm 2.4 requires a rather straightforward manipulation of the definition (4)
of the derivative. Parts LI and h should he proved as an illustrative exercise. The
[2]
20 ONE“ARlABLE CALCULUS: FOUNDATIONS
Theorem 2 . 4 Suppose that k is an arbitrary constant and that f and g are
differentiable functions at x = x0, Then,
4 (f 2 d™(X”) = f™(xo) 2 g™(˜ob
= k(f ˜(.4)>
b) (kf )˜@o)
proofs of parts c, d, and e are a little more subtle. The proof of part f is listed as
an exercise below for negative integers k and will be carried out for fractions k in
Section 4.2.
Example 2.7 We use Theorems 2.3 and 2.4 to calculate the derivatives of some
simple functions.
a) (2 + 3x”  4x™ + 5)™ = 7x” + 18x5 xx,
b) ((.r™ + 3 x  1)(.x™  8.˜))™ = (2x + 3)(x irn)
+ (x™ + 3x  1)(4x™  8)
e) ((2 4x™ + I)˜)™ = 5(x3 4x™ + 1)“. (32 Xx):
yl/™
= 3x˜,
f) (3x”? + 3x ˜I)™
L2.51 “IFFERENTIABIL,TYANDCONTIN”ITY 2 9
EXERCISES
2.10 a) Use the geometric definition of the derivative to prove that the derivative of a
constant function is 0 everywhere and the derivative of f(x) = mx is f™(x) = m
for all x.
b) Use the method of the proof of Theorem 2.2 to prove that the derivative of x3 is
3x™ and the derivative of x™ is 4x1.
2.11 Find the derivative of the following funclions at an arbitrary point:
b ) IX™,
a )  72,