Math 31A Discussion Notes
Week 3
October 13 and October 15, 2015
This week in lecture you’re discussing three important rules for computing derivatives:
the product rule, the quotient rule, and the chain rule. In discussion we’ll work a few
examples from topics leading up to these rules, and then review these rules, time permitting.
Examples
One of your homework questions this week asks you to find the horizontal and vertical
asymptotes of some functions. Recall that the line y = L is called a horizontal asymptote
of y = f (x) if limx→∞ f (x) = L or if limx→−∞ f (x) = L, and the line x = L is called a
vertical asymptote of y = f (x) if limx→L− = ±∞ or if limx→L+ = ±∞.
√
√
Example. Find the horizontal and vertical asymptotes of f (x) = x2 + 2x + 4− x2 + 2x,
if there are any.
(Solution) We begin by checking for a horizontal asymptote. Note that we cannot just split
the limit over subtraction:
√
√
√
√
2
2
2
2
x + 2x + 4 − x + 2x 6= lim
x + 2x + 4 − lim
x + 2x .
lim
x→∞
x→∞
x→∞
Neither of the limits on the right exists, so this wouldn’t make sense. What we can do,
however, is rewrite our limit as a quotient:
√
√
lim
x2 + 2x + 4 − x2 + 2x
x→∞
!
√
√
2 + 2x + 4 +
2 + 2x
√
√
x
x
√
= lim ( x2 + 2x + 4 − x2 + 2x) · √
x→∞
x2 + 2x + 4 + x2 + 2x
x2 + 2x + 4 − (x2 + 2x)
√
= lim √
x→∞
x2 + 2x + 4 + x2 + 2x
4
√
= lim √
= 0.
x→∞
x2 + 2x + 4 + x2 + 2x
So f (x) has a horizontal asymptote of y = 0. To check for vertical asymptotes of f (x), we
first need to find values where an infinite limit might occur. Since f is continuous on its
domain, we begin by determining which values are missing from the domain of f . A value
will be missing from the domain of f when either of the involved square roots cannot be
found. Note that x2 + 2x + 4 = (x + 1)2 + 3 ≥ 3, so the first square root is always defined.
For the second square root, however, we notice that x2 + 2x = x(x + 2), so this square root
is not defined when −2 < x < 0. So f (x) is not defined when −2 < x < 0, meaning that we
should check the following two limits for vertical asymptotes:
lim f (x)
x→−2−
and
1
lim f (x).
x→0+
But f is continuous on its domain, and both −2 and 0 are in the domain of f , so
p
p
√
√
lim− f (x) = (−2)2 + 2(−2) + 4 − (−2)2 + 2(−2) = 4 − 0 = 2
x→2
and
lim+ f (x) =
p
(0)2 + 2(0) + 4 −
p
√
√
(0)2 + 2(0) = 4 − 0 = 2.
x→0
So f (x) has no vertical asymptotes.
Before we can compute derivatives with our new rules, we first must be able to compute
them from the limit definition, as in the following two examples.
Example.
1. Show that f 0 (0) does not exist, given that f (x) = x1/3 .
(Solution) If f 0 (0) exists, it is equal to limh→0
f (0+h)−f (0)
,
h
so we compute this limit:
f (0 + h) − f (0)
(0 + h)1/3 − 01/3
= lim
h→0
h→0
h
h
1
h1/3
= lim 2/3 .
= lim
h→0 h
h→0 h
lim
But the denominator of this last ratio tends to 0 while the numerator is constant. So
the fraction will become arbitrarily large as h → 0, and we conclude that this limit
does not exist. If we look at the graph of f (x), it’s not surprising that this limit grows
without bound; the secant lines passing through (0, 0) and some nearby point (h, f (h))
become arbitrarily steep as h approaches 0.
2. Let g(x) = x4 and compute g 0 (x) from the limit definition.
(Solution) We have
(x + h)4 − x4
g(x + h) − g(x)
= lim
h→0
h→0
h
h
x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 − x4
= lim
h→0
h
3
2 2
4x h + 6x h + 4xh3 + h4
= lim
h→0
h
= lim (4x3 + 6x2 h + 4xh2 + h3 ) = 4x3 ,
lim
h→0
so g 0 (x) = 4x3 , for all x ∈ R.
2
Product and Quotient Rules
In class you proved the product rule,
d
(f (x)g(x)) = f 0 (x)g(x) + f (x)g 0 (x),
dx
and the quotient rule,
d
dx
f (x)
g(x)
=
f 0 (x)g(x) − f (x)g 0 (x)
.
(g(x))2
Here we’ll practice using them1 .
Example.
1. Use the product rule to compute
(Solution) Since
d
((x2 − 1)(3x + 3)).
dx
d
d 2
(x − 1) = 2x and
(3x + 3) = 3, we have
dx
dx
d
(x2 − 1)(3x + 3) = (2x)(3x + 3) + (x2 − 1)(3) = 9x2 + 6x − 3.
dx
2. Use the product rule to compute
(Solution) Since
d
((3x + 4)2 ).
dx
d
(3x + 4) = 3, we have
dx
d
d
(3x + 4)2 =
((3x + 4)(3x + 4)) = (3)(3x + 4) + (3x + 4)(3)
dx
dx
= 6(3x + 4) = 18x + 24.
d
3. Use the quotient rule to compute
dx
(Solution) We have
Equivalently,
d 1
( )
dx x
1
.
x
d
d
(1) = 0 and
(x) = 1, so
dx
dx
(0)(x) − (1)(1)
−1
d 1
=
= 2.
2
dx x
x
x
=
d
(x−1 )
dx
= −x−2 =
1
−1
.
x2
Once we know the chain rule, the quotient rule is an immediate consequence of the product rule. (Prove
it!) But when you proved these in class, you didn’t have the chain rule yet, and thus couldn’t use it to prove
the quotient rule.
3
d
4. Use the quotient rule to compute
dx
3
.
x6
d 6
d
(3) = 0 and
(x ) = 6x5 , so
dx
dx
d
3
(0)(x6 ) − (3)(6x5 )
−18x5
−18
=
=
=
.
dx x6
(x6 )2
x12
x7
(Solution) We have
Equivalently,
d 3
( )
dx x6
=
d
(3x−6 )
dx
= −18x−7 =
d
5. Use the quotient rule to compute
dx
−18
.
x7
2x2 − 7x + 3
.
4x − 7
d
d
(2x2 − 7x + 3) = 4x − 7 and
(4x − 7) = 4, so
dx
dx
d 2x2 − 7x + 3
(4x − 7)(4x − 7) − (2x2 − 7x + 3)(4)
=
dx
4x − 7
(4x − 7)2
8x2 − 28x + 37
.
=
16x2 − 56x + 49
(Solution) We have
Chain Rule
Another rule you proved in class is the chain rule:
d
((f ◦ g)(x)) = f 0 (g(x)) · g 0 (x).
dx
The following examples give us practice with using the chain rule.
Example.
1. Use the chain rule to compute
d
((3x + 4)2 ).
dx
(Solution) We can write (3x + 4)2 = f (g(x)), where g(x) = 3x + 4 and f (x) = x2 .
Notice that g 0 (x) = 3 and f 0 (x) = 2x, so
d
(3x + 4)2 = f 0 (g(x)) · g 0 (x) = 2(3x + 4) · 3 = 18x + 24.
dx
d
Note that this agrees with our computation of
((3x + 4)2 ) done via the product
dx
rule.
4
d
2. Use the chain rule to compute
dx
(Solution) Note that
and g 0 (x) = 6x5 , so
3
x6
d
dx
3
.
x6
= 3f (g(x)), where f (x) =
3
x6
d
= 3 (f (g(x))) = 3
dx
−6x5
−18
= 3 12 = 7 .
x
x
1
x
and g(x) = x6 . Then f 0 (x) =
−1
(x6 )2
−1
x2
(6x5 )
Again, this agrees with our computation that used the quotient rule.
3. Use the chain rule to compute
d √ 2
x + 2x + 4 .
dx
√
√
(Solution) Write x2 + 2x + 4 = f (g(x)), where f (x) = x and g(x) = x2 + 2x + 4.
Then, since f 0 (x) = 2√1 x and g 0 (x) = 2x + 2,
d √ 2
x + 2x + 4 =
dx
1
√
2
2 x + 2x + 4
x+1
(2x + 2) = √
.
2
x + 2x + 4
d
4. Use the chain rule and the power rule for positive integers to compute
xp/q , where
dx
p and q are positive integers.
dy
d
(Solution) First, define a new variable y = xp/q . Notice that dx
= dx
(xp/q ), and that
q
p
y = x . We can take the derivative with respect to x on either side of the latter
equation to find that
dy
= pxp−1 .
qy q−1
dx
(We use the chain rule to take the derivative on the left.) Dividing both sides by qy q−1 ,
we have
dy
p(1−q)
d
p
p
p p−q
p p
xp/q =
= xp−1 y 1−q = xp−1 x q = x q = x q −1 .
dx
dx
q
q
q
q
So the power rule we like to use for positive integers extends to positive rational
numbers.
5