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The Derivative Limits Onesided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules:
The HigherOrder Derivatives The Chain Rule Differentiation of Exponential and
Differentiation
Logarithmic Functions
Logarithmic Marginal Functions in Economics 9.1
9.1
Limits
y
400 f ( x) 300
200
100
10 20 30 40 50 60 x Introduction to Calculus
Introduction Historically, the development of calculus by Isaac Newton
Historically,
Isaac and Gottfried W. Leibniz resulted from the investigation
Gottfried
of the following problems:
of
1. Finding the tangent line to a curve at a given point on the
Finding
tangent
curve:
curve:
y
T t Introduction to Calculus
Introduction Historically, the development of calculus by Isaac Newton
Historically,
Isaac and Gottfried W. Leibniz resulted from the investigation
Gottfried
of the following problems:
of
1. Finding the area of a planar region bounded by an
Finding
area
arbitrary curve.
arbitrary
y R t Introduction to Calculus
Introduction The study of the tangentline problem led to the
The
tangentline creation of differential calculus, which relies on the
differential
which
concept of the derivative of a function.
derivative The study of the area problem led to the creation of
The
area
integral calculus, which relies on the concept of the
integral
which
antiderivative, or integral, of a function.
antiderivative or integral Example: A Speeding Maglev
Example: From data obtained in a test run conducted on a
From
test prototype of maglev, which moves along a straight
prototype
which
monorail track, engineers have determined that the
position of the maglev (in feet) from the origin at time t
position
from
time
is given by
is
s = f(t) = 4t2
(0 ≤ t ≤ 30)
30) f is called the position function of the maglev.
position The position of the maglev at time t = 0, 1, 2, 3, … , 10 is
0,
The position f(0) = 0
(0) f(1) = 4
(1) f(2) = 16
(2) f(3) = 36 … f(10) = 400
(3) But what if we want to find the velocity of the maglev at
But
velocity any given point in time?
point Example: A Speeding Maglev
Example: Say we want to find the velocity of the maglev at t = 2.
Say
velocity We may compute the average velocity of the maglev
We
average over an interval of time, such as [2, 4] as follows:
interval
such [2,
Distance covered f (4) − f (2)
=
Time elapsed
4−2
4(42 ) − 4(22 )
=
2
64 − 16
=
2
= 24 or 24 feet/second.
or 24 This is not the velocity of the maglev at exactly t = 2,
This not
exactly
but it is a useful approximation.
approximation Example: A Speeding Maglev
Example: We can find a better approximation by choosing a
We
better smaller interval to compute the speed, say [2, 3].
smaller
[2, More generally, let t > 2. Then, the average velocity of
let
Then,
average
the maglev over the time interval [2, t] is given by
time
Distance covered f (t ) − f (2)
=
Time elapsed
t−2
4(t 2 ) − 4(22 )
=
t−2
4(t 2 − 4)
=
t−2 Example: A Speeding Maglev
Example:
4(t 2 − 4)
Average velocity =
t−2 By choosing the values of t closer and closer to 2, we
By
closer
we
obtain average velocities of the maglev over smaller and
average
smaller time intervals.
smaller The smaller the time interval, the closer the average
The smaller
time
the closer
velocity becomes to the instantaneous velocity of the train
instantaneous
at t = 2, as the table below demonstrates:
t 2.5 2.1 2.01 2.001 2.0001 Average Velocity 18 16.4 16.04 16.004 16.0004 The closer t gets to 2, the closer the average velocity gets
The closer
the closer
average to 16 feet/second.
16 Thus, the instantaneous velocity at t = 2 seems to be
Thus,
instantaneous
16 feet/second.
16 Intuitive Definition of a Limit
Intuitive Consider the function g, which gives the average velocity
Consider
g,
average of the maglev:
of 4(t 2 − 4)
g (t ) =
t −2 Suppose we want to find the value that g(t) approaches
Suppose
approaches
as t approaches 2.
approaches
✦ We take values of t approaching 2 from the right
We
approaching from
(as we did before), and we find that g(t) approaches
approaches
16:
16
t 2.5 2.1 2.01 2.001 2.0001 g (t ) 18 16.4 16.04 16.004 16.0004 ✦ Similarly, we take values of t approaching 2 from the left,
Similarly,
approaching from
and we find that g(t) also approaches 16:
approaches 16
t 1.5 1.9 1.99 1.999 1.9999 g(t) 14 15.6 15.96 15.996 15.9996 Intuitive Definition of a Limit
Intuitive We have found that as t approaches 2 from either side,
We
approaches from g(t) approaches 16.
approaches 16 In this situation, we say that the limit of g(t) as t
In
the
approaches 2 is 16.
approaches
16 This is written as
4(t 2 − 4)
lim g (t ) = lim
= 16
t →2
t →2
t −2 Observe that t = 2 is not in the domain of g(t) .
Observe
not But this does not matter, since t = 2 does not play any
But
does
since role in computing this limit.
role Limit of a Function
Limit The function f has a limit L as x approaches a, written
The function
limit
approaches lim f ( x ) = L
x→a
if the value of f(x) can be made as close to the number
if
close
L as we please by taking x values sufficiently close to
values sufficiently
(but not equal to) a.
not Examples
Examples Let f(x) = x3. Evaluate
Let
Evaluate lim f ( x ).
x→2 Solution
Solution You can see in the graph
You
that f(x) can be as close
can
close
to 8 as we please by taking
x sufficiently close to 2.
sufficiently Therefore,
Therefore, f( x ) = x y
8 3 6
4
2 lim x 3 = 8
x→2 –2 –1 1
–2 2 3 x Examples
Examples
x + 2
1 Let g ( x ) = Let if x ≠ 1
if x = 1 Evaluate lim g ( x ).
x→1 Solution
Solution You can see in the graph that
You
g(x) can be as close
can
close
to 3 as we please by taking
x sufficiently close to 1.
sufficiently Therefore,
Therefore, y g(x) 5
3
1 lim g ( x ) = 3
x→1 –2 –1 1 2 3 x Examples
Examples Let f ( x ) =
Let 1
x2 Evaluate lim f ( x ). Solution
Solution The graph shows us that as x
The
approaches 0 from either
approaches
side, f(x) increases without
side
bound and thus does not
bound
not
approach any specific real
number.
number. Thus, the limit of f(x) does
Thus,
not exist as x approaches 0.
not
approaches x→0 y
5 f ( x) = –2 –1 1 2 1
x2 x Theorem 1
Theorem
Properties of Limits lim f ( x ) = L
x→a Suppose
Then,
1. lim [
x→a and lim g ( x ) = M x→a r f ( x )] = lim f ( x ) = Lr x →a r r, a real number 2. lim cf ( x ) = c lim f ( x ) = cL
2. x→a
x →a
3. lim
x →a [ f ( x ) ± g ( x )] = lim
x→a 4. lim
4. x→a [ f ( x ) g ( x )] = lim x→a 5.
5. f ( x ) ± lim g ( x ) = L ± M
x→a f ( x ) lim g ( x ) = LM x→a f ( x ) lim f ( x ) L
lim
= x→a
=
x →a g ( x )
lim g ( x ) M
x→a c, a real number Provided that M ≠ 0
Provided Examples
Examples Use theorem 1 to evaluate the following limits:
Use theorem
3 lim x = lim x = 23 = 8
x→2
3 x→2 3/2 lim 5x = 5 lim x = 5(4)3/2 = 40
x→4
3/2 x→4 4 lim (5x − 2) = 5 lim x − lim 2 = 5(1)4 − 2 = 3
x→1
4 x→1 x→1 Examples
Examples Use theorem 1 to evaluate the following limits:
Use theorem lim 2x
x →3 x + 7 = 2 lim x x→3 3 2 3 lim x 2 + 7 = 2(3)3 (3)2 + 7 = 216
x→3 lim (2 x 2 + 1) 2(2)2 + 1 9
2x + 1
= x→2
=
= =3
lim
x→2
lim ( x + 1)
2 +1
3
x +1
x→2
2 Indeterminate Forms
Indeterminate
4( x 2 − 4) Let’s consider lim
x →2 x − 2
which we evaluated earlier for the maglev example by
which
maglev
looking at values for x near x = 2.
near If we attempt to evaluate this expression by applying
If
Property 5 of limits, we get
Property lim 4( x 2 − 4) 0
4( x − 4) x→2
lim
=
=
x →2 x − 2
lim x − 2
0
x →2
2 In this case we say that the limit of the quotient f(x)/g(x) as
In
quotient x approaches 2 has the indeterminate form 0/0.
approaches
0/0 This expression does not provide us with a solution to our
This
not
solution
problem.
problem. Strategy for Evaluating Indeterminate Forms
Strategy 1. Replace the given function with an appropriate one that takes on the same values as the
same
original function everywhere except at x = a.
everywhere except
2. Evaluate the limit of this function as x
approaches a.
approaches Examples
Examples
4( x 2 − 4) Evaluate lim
x →2 x − 2 Solution As we’ve seen, here we have an indeterminate form 0/0.
As
indeterminate
0/0 We can rewrite 4( x − 2)( x + 2)
We 4( x 2 − 4)
=
= 4( x + 2)
x−2
x−2
x≠2 Thus, we can x 2 − 4)
4( say that lim
x →2 x−2 = lim 4( x + 2) = 16
x →2 Note that 16 is the same value we obtained for the maglev
Note
16 example through approximation.
example
approximation Examples
Examples
4( x 2 − 4) Evaluate lim
x →2 x − 2 Solution Notice in the graphs below that the two functions yield the
Notice
same graphs, except for the value x = 2:
same
except
4( x 2 − 4)
f ( x) =
x−2 y
20 20 16 8 4
–1 12 8 –2 16 12 –3 g ( x ) = 4( x + 2) y 4
1 2 3 x –3 –2 –1 1 2 3 x Examples
Examples Evaluate 1+ h −1
lim
h→0
h Solution As we’ve seen, here we have an indeterminate form 0/0.
As
indeterminate
0/0 We can rewrite (with the constraint that h ≠ 0):
We
constraint
1+ h −1
1+ h −1 1+ h +1
h
1
=
⋅
=
=
h
h
1 + h + 1 h( 1 + h + 1)
1+ h +1 Thus, we can say that 1+ h −1
1
1
1
lim
= lim
=
=
h→0
h→0 1 + h + 1
h
1 +1 2 Limits at Infinity
Limits There are occasions when we want to know whether f(x)
There approaches a unique number as x increases without
approaches unique
bound.
bound In the graph below, as x increases without bound, f(x)
In
without
approaches the number 400.
400 We call the line y = 400
We
y
a horizontal asymptote.
horizontal
400 In this case, we can say
In
f ( x)
that
that
300
lim f ( x ) = 400
x→∞ and we call this a limit
and
limit
of a function at infinity. 200
100 10 20 30 40 50 60 x Example
Example
2x2 Consider the function f ( x ) =
1 + x2 Determine what happens to f(x) as x gets larger and larger.
Determine
larger
Solution We can pick a sequence of values of x and substitute them
We
sequence in the function to obtain the following values:
in
x 1 2 5 10 100 1000 f(x) 1 1.6 1.92 1.98 1.9998 1.999998 As x gets larger and larger, f(x) gets closer and closer to 2.
As
larger
closer Thus, we can say that
2x2 lim x→∞ 1 + x 2 =2 Limit of a Function at Infinity
Limit The function f has the limit L as x increases without
The
limit bound (as x approaches infinity), written
bound
approaches lim f ( x ) = L x→∞ if f(x) can be made arbitrarily close to L by taking x
if
close
large enough.
large Similarly, the function f has the limit M as x decreases
Similarly,
limit
without bound (as x approaches negative infinity),
without
approaches
),
written
written
lim f ( x ) = M
x→−∞ if f(x) can be made arbitrarily close to M by taking x
if
close
large enough in absolute value.
large Examples
Examples −1 Let f ( x ) = Let
1 Evaluate if x < 0
if x ≥ 0 lim f ( x ) lim f ( x ) and
x→∞ x→−∞ Solution Graphing f(x) reveals that
Graphing y lim f ( x ) = 1 x→∞ lim f ( x ) = −1
x→−∞ f ( x) 1
–3 3
–1 x Examples
Examples Let g ( x ) =
Let Evaluate 1
x2 lim g ( x ) and lim g ( x ) x→∞ x→−∞ Solution Graphing g(x) reveals that
Graphing y 1
g ( x) = 2
x lim g ( x ) = 0
x→∞
lim g ( x ) = 0
x→−∞
–3 –2 –1 1 2 3 x Theorem 2
Theorem
Properties of Limits All properties of limits listed in Theorem 1 are valid when a is replaced by ∞ or –∞ . In addition, we have the following properties for
limits to infinity: For all n > 0,
For 1
lim
=0
x→∞ x n and 1
provided that n is defined.
x 1
lim
=0
x→−∞ x n Examples
Examples
x2 − x + 3 Evaluate lim
Evaluate x→∞
2 x3 + 1 Solution
Solution The limits of both the numerator and denominator do not
The limits
numerator
denominator
exist as x approaches infinity, so property 5 is not
exist
property
not
applicable.
applicable. We can find the solution instead by dividing numerator
We
dividing
and denominator by x3:
and
by
11
3
− 2+ 3
( x 2 − x + 3) / x 3
xx
x = 0−0+0 = 0 = 0
lim
= lim
x→∞ (2 x 3 + 1) / x 3
x→∞
1
2+0
2
2+ 3
x Examples
Examples
3x 2 + 8 x − 4 Evaluate lim
Evaluate x→∞
2 x2 + 4 x − 5 Solution
Solution Again, we see that property 5 does not apply.
Again,
property
not So we divide numerator and denominator by x2:
So
divide
and
by
84
−2
(3x + 8 x − 4) / x
x x = 3+ 0−0 = 3
lim 2
= lim
x→∞ (2 x + 4 x − 5) / x 2
x→∞
45
2+ − 2 2+0−0 2
xx
2 2 3+ Examples
Examples
2 x 3 − 3x 2 + 1 Evaluate lim
Evaluate x→∞ 2
x + 2x + 4 Solution
Solution Again, we see that property 5 does not apply.
Again,
property
not But dividing numerator and denominator by x2 does not
But dividing
and
by
not
1
help in this case:
help
2x − 3 + 2
3
2
2 lim
x→∞ (2 x − 3x + 1) / x
x
= lim
x→∞
24
( x 2 + 2 x + 4) / x 2
1+ + 2
xx In other words, the limit does not exist.
In
the We indicate this by writing 2
3 lim
x→∞ 2 x − 3x + 1
=∞
2
x + 2x + 4 9.2
9.2
OneSided Limits and Continuity y a b c d x OneSided Limits
OneSided
x −1
x +1 Consider the function f ( x ) = if x < 0
if x ≥ 0 Its graph shows that
Its f does not have a
does not
limit as x approaches
limit
zero, because
zero because
approaching from each
side results in different
values.
values y f ( x)
1
–1 1
–1 x OneSided Limits
OneSided
x −1
x +1 Consider the function f ( x ) = if x < 0
if x ≥ 0 If we restrict x to be greater
If
restrict
greater than zero (to the right of zero),
than
we see that f(x) approaches 1 as
approaches
close as we please as x
approaches 0.
approaches In this case we say that the
In
righthand limit of f as x
righthand
approaches 0 is 1, written
approaches lim f ( x ) = 1 x→0+ y f ( x)
1
–1 1
–1 x OneSided Limits
OneSided
x −1
x +1 Consider the function f ( x ) = if x < 0
if x ≥ 0 Similarly, if we restrict x to be
Similarly,
restrict less than zero (to the left of zero),
less
we see that f(x) approaches –1
approaches –1
as close as we please as
x approaches 0.
approaches In this case we say that the
In
lefthand limit of f as x
lefthand
approaches 0 is – 1, written
approaches lim f ( x ) = −1 x→0− y f ( x)
1
–1 1
–1 x OneSided Limits
OneSided The function f has the righthand limit L as x
The
righthand approaches from the right, written
approaches lim f ( x ) = L x→a + if the values of f(x) can be made as close to L as we
if
please by taking x sufficiently close to (but not equal
to) a and to the right of a.
to Similarly, the function f has the lefthand limit L as x
Similarly,
lefthand
approaches from the left, written
approaches lim f ( x ) = L x→a − if the values of f(x) can be made as close to L as we
if
please by taking x sufficiently close to (but not equal
to) a and to the left of a.
to Theorem 3
Theorem
Properties of Limits The connection between onesided limit and the twosided limit defined earlier is given by the following theorem. Let f be a function that is defined for all values of x
Let close to x = a with the possible exception of a itself.
Then
Then lim f ( x ) = L
x →a if and only if lim f ( x ) = lim f ( x ) = L x→a − x→a + Examples Show that lim f ( x ) exists by studying the onesided
Show
onesided
x →0 limits of f as x approaches 0:
limits
approaches
x f ( x) = −x Solution
Solution For x > 0, we find
For if x > 0
if x ≤ 0
y lim f ( x ) = 0 x→0+ And for x ≤ 0, we find
And lim f ( x ) = 0 1 x→0− Thus,
Thus, lim f ( x ) = 0
x →0 f ( x) 2 –2 –1 1 2 x Examples
Examples Show that lim g ( x ) does not exist.
Show
x →0 −1
g ( x) = 1 if x < 0
if x ≥ 0 Solution For x < 0, we find
For y lim g ( x) = −1 x →0− And for x ≥ 0, we find
And lim g ( x) = 1 g ( x)
1 x →0+ Thus,
Thus, x lim g ( x) does
x →0 not exist.
not –1 Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y a x This function is discontinuous at the following points:
This
discontinuous ✦ At x = a, f is not defined (x = a is not in the domain of f ).
At
not
not
). Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y a b x This function is discontinuous at the following points:
This
discontinuous ✦ At x = b, f(b) is not equal to the limit of f(x) as x approaches b.
At
not
of
as approaches Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y b c x This function is discontinuous at the following points:
This
discontinuous ✦ At x = c, the function does not have a limit, since the lefthand
At
does
since
lefthand
and righthand limits are not equal.
righthand
not Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y c d x This function is discontinuous at the following points:
This
discontinuous ✦ At x = d, the limit of the function does not exist, resulting in a
At
limit
resulting
break in the graph.
break Continuity of a Function at a Number
Continuity A function f is continuous at a number x = a
function
continuous if the following conditions are satisfied:
conditions
1. f(a) is defined.
2. lim f ( x ) exists.
x→a
3. lim f ( x ) = f (a )
3. x→a If f is not continuous at x = a, then f is said to
If
then be discontinuous at x = a.
discontinuous Also, f is continuous on an interval if f is
Also,
continuous
continuous at every number in the interval.
continuous Examples
Examples Find the values of x for which the function is continuous:
Find
continuous f ( x) = x + 2 Solution The function f is continuous everywhere because the three
The
continuous
conditions for continuity are satisfied for all values of x.
y
5 f ( x) = x + 2 4
3
2 1
– 2 –1 1 2 x Examples
Examples Find the values of x for which the function is continuous:
Find
continuous x2 − 4
g ( x) =
x−2 Solution The function g is discontinuous at x = 2 because g is not
The
defined at that number. It is continuous everywhere else.
defined
y x2 − 4
g ( x) =
x−2 5
4
3
2 1
– 2 –1 1 2 x Examples
Examples Find the values of x for which the function is continuous:
Find
continuous x + 2
h( x ) = 1 if x ≠ 2
if x = 2 Solution The function h is continuous everywhere except at x = 2
The
continuous
except
where it is discontinuous because
because
h (2) = 1 ≠ lim h ( x ) = 4
x→2 5 y y = h( x ) 4
3
2 1
– 2 –1 1 2 x Examples
Examples Find the values of x for which the function is continuous:
Find
continuous −1 if x < 0
F ( x) = 1 if x ≥ 0 Solution The function F is discontinuous at x = 0 because the limit
The
at
of F fails to exist as x approaches 0. It is continuous
fails
approaches It
everywhere else.
y y = F ( x) 1 x
–1 Examples
Examples Find the values of x for which the function is continuous:
Find
continuous 1
if x > 0 G( x) = x −1 i f x ≤ 0 Solution The function G is discontinuous at x = 0 because the limit
The
at
of G fails to exist as x approaches 0. It is continuous
fails
approaches It
everywhere else.
y y = G ( x)
x
–1 Properties of Continuous Functions
Properties
1. The constant function f(x) = c is continuous everywhere.
The constant
continuous
2. The identity function f(x) = x is continuous everywhere.
The identity
continuous
If f and g are continuous at x = a, then
and are
at
1. [f(x)]n, where n is a real number, is continuous at
where
continuous
x = a whenever it is defined at that number.
2. f ± g is continuous at x = a.
continuous
3. fg is continuous at x = a.
fg continuous
4. f /g is continuous if g(a) ≠ 0.
/g continuous Properties of Continuous Functions
Properties Using these properties, we can obtain the following
Using additional properties.
additional 1. A polynomial function y = P(x) is continuous at every
polynomial
value of x.
2. A rational function R(x) = p(x)/q(x) is continuous at
rational
every value of x where q(x) ≠ 0. Examples
Examples Find the values of x for which the function is continuous.
Find
values
continuous f ( x ) = 3x 3 + 2 x 2 − x + 10 Solution The function f is a polynomial function of degree 3, so f(x)
The
polynomial
so
is continuous for all values of x.
continuous Examples
Examples Find the values of x for which the function is continuous.
Find
values
continuous 8 x10 − 4 x 2 + 1
g ( x) =
x2 + 1 Solution The function g is a rational function.
The
rational Observe that the denominator of g is never equal to zero.
Observe
denominator
never Therefore, we conclude that g(x) is continuous for all
Therefore,
values of x.
values Examples
Examples Find the values of x for which the function is continuous.
Find
values
continuous 4 x 3 − 3x 2 + 1
h( x ) = 2
x − 3x + 2 Solution The function h is a rational function.
The
rational In this case, however, the denominator of h is equal to zero
In
denominator
equal
at x = 1 and x = 2, which we can see by factoring. Therefore, we conclude that h(x) is continuous everywhere
Therefore,
except at x = 1 and x = 2. Intermediate Value Theorem
Intermediate Let’s look again at the maglev example.
Let’s
maglev The train cannot vanish at any instant of time and cannot
The
cannot skip portions of track and reappear elsewhere.
skip Intermediate Value Theorem
Intermediate Mathematically, recall that the position of the maglev is a
recall function of time given by f(t) = 4t2 for 0 ≤ t ≤ 30:
for
y y = 4t 2 s2
s3
s1 t1 t3 t2 t Suppose the position of the maglev is s1 at some time t1 and its
Suppose
position
time position is s2 at some time t2.
position
time Then, if s3 is any number between s1 and s2, there must be at
Then,
is
least one t3 between t1 and t2 giving the time at which the
least
maglev is at s3 (f(t3) = s3). Theorem 4
Theorem
Intermediate Value Theorem
Intermediate The Maglev example carries the gist of the
The intermediate value theorem:
intermediate If f is a continuous function on a closed interval [a, b]
If
continuous
closed and M is any number between f(a) and f(b), then there
is at least one number c in [a, b] such that f(c) = M.
is
y y
f(b) f(b) y = f ( x) y = f ( x)
M
f(a) M
f(a) a c b x a c1 c2 c3 b x Theorem 5
Existence of Zeros of a Continuous Function
Existence A special case of this theorem is when a continuous
special function crosses the x axis.
function
axis If f is a continuous function on a closed interval [a, b],
If
continuous
closed and if f(a) and f(b) have opposite signs, then there is at
least one solution of the equation f(x) = 0 in the
least
interval (a, b).
y y
f(b) f(b) y = f ( x) y = f ( x) a
f(a) c b x a c1
f(a) c2 c3 b x Example
Example Let f(x) = x3 + x + 1.
Let a. Show that f is continuous for all values of x.
Show
continuous
b. Compute f(–1) and f(1) and use the results to deduce that
Compute
there must be at least one number x = c, where c lies in the
where
interval (–1, 1) and f(c) = 0.
(–1,
Solution
a. The function f is a polynomial function of degree 3 and is
The
polynomial
therefore continuous everywhere.
continuous
b. f (–1) = (–1)3 + (–1) + 1 = –1 and f (1) = (1)3 + (1) + 1 = 3
(–1)
and (1)
Since f (–1) and f (1) have opposite signs, Theorem 5
and
opposite
Theorem
tells us that there must be at least one number x = c
with
–1 < c < 1 such that f(c) = 0.
–1 9.3
9.3
The Derivative
y
f(x + h)
f(x + h) – f(x)
f ( x) A
h x x+h x An Intuitive Example
An Consider the maglev example from Section 2.4.
Consider
maglev The position of the maglev is a function of time given by
The position
time (0 ≤ t ≤ 30)
(0 s = f(t) = 4t2 where s is measured in feet and t in seconds.
where
feet
seconds Its graph is: s (ft)
(ft) s = f (t ) 60
40
20
1 2 3 4 t (sec)
(sec) An Intuitive Example
An The graph rises slowly at first but more rapidly over time.
The
slowly at
more
over This suggests the steepness of f(t) is related to the speed of
This
steepness
speed the maglev, which also increases over time.
over If so, we might be able to find the speed of the maglev at
If
find
speed
any given time by finding the steepness of f at that time.
steepness
at But how do we find the steepness of a point at a curve?
s (ft)
(ft) s = f (t ) 60
40
20
1 2 3 4 t (sec)
(sec) Slopes of Lines and of Curves The slope at a point of a curve is given by the slope of the
The slope
curve
slope tangent to the curve at that point:
tangent y A Suppose we want to find the
Suppose
slope at point A.
slope
The tangent line has the same
The tangent
slope as the curve does at
slope
curve
point A. x Slopes of Lines and of Curves The slope at a point of a curve is given by the slope of the
The slope
curve
slope tangent to the curve at that point:
tangent y Slope = 1.8 ∆ y = 1.8
1.8
A
∆x = 1 The slope of the tangent in this
The slope
case is 1.8:
1.8 ∆y 1.8
Slope =
=
= 1.8
∆x
1
x Slopes of Lines and of Curves The slope at a point of a curve is given by the slope of the
The slope
curve
slope tangent to the curve at that point:
tangent y Slope = 1.8 A x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we
To
accurately
tangent
curve we must make the change in x as small as possible:
change
small as y Slope = 1.8 Slope = ∆y 3
= = 0.75
∆x 4 ∆y = 3
A As we let ∆ x get smaller, the
As
smaller the
slope of the secant becomes
slope of
secant
closer to the slope of the
closer
of
tangent to the curve at that
tangent
point.
point. ∆x = 4 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we
To
accurately
tangent
curve we must make the change in x as small as possible:
change
small as y Slope = 1.8 Slope = ∆y 2.4
=
= 0.8
∆x
3 ∆ y = 2.4
2.4
A As we let ∆ x get smaller, the
As
smaller the
slope of the secant becomes
of
secant
closer to the slope of the
closer
of
tangent to the curve at that
tangent
point.
point. ∆x = 3 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we
To
accurately
tangent
curve we must make the change in x as small as possible:
change
small as y Slope = 1.8 Slope = ∆ y = 2.2
2.2
A ∆y 2.2
=
= 1.1
∆x
2 As we let ∆ x get smaller, the
As
smaller the
slope of the secant becomes
of
secant
closer to the slope of the
closer
of
tangent to the curve at that
tangent
point.
point. ∆x = 2 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we
To
accurately
tangent
curve we must make the change in x as small as possible:
change
small as y Slope = 1.8 A Slope = ∆y 1.5
=
= 1.5
∆x
1 As we let ∆ x get smaller, the
As
smaller the
slope of the secant becomes
of
secant
closer to the slope of the
closer
of
tangent to the curve at that
tangent
point.
point. ∆ y = 1.5
1.5
∆x = 1 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we
To
accurately
tangent
curve we must make the change in x as small as possible:
change
small as y Slope = 1.8 Slope = ∆y 0.00179
=
≈ 1.8
∆x
0.001 As we let ∆ x get smaller, the
As
smaller the
slope of the secant becomes
of
secant
closer to the slope of the
closer
of
tangent to the curve at that
tangent
point.
point. A ∆ y = 0.00179
0.00179
∆ x = 0.001
0.001 x Slopes of Lines and of Curves In general, we can express the slope of the secant as follows:
In
of
Slope = ∆y f ( x + h ) − f ( x ) f ( x + h ) − f ( x )
=
=
∆x
( x + h) − x
h y
f(x + h)
f(x + h) – f(x) f (x ) A
h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches
Thus,
approaches
the slope of
secant approaches the slope of the tangent to the curve at that point:
slope of
tangent
y
f(x + h)
f(x + h) – f(x) f (x ) A
h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches
Thus,
approaches
the slope of
secant approaches the slope of the tangent to the curve at that point:
slope of
tangent
y
f(x + h)
f(x + h) – f(x)
f( x ) A
h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches
Thus,
approaches
the slope of
secant approaches the slope of the tangent to the curve at that point:
slope of
tangent
y
f(x + h)
f(x + h) – f(x)
f( x ) A
h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches
Thus,
approaches
the slope of
secant approaches the slope of the tangent to the curve at that point:
slope of
tangent
y f(x + h)
f( x ) f(x + h) – f(x) A
h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches
Thus,
approaches
the slope of
secant approaches the slope of the tangent to the curve at that point:
slope of
tangent
y f(x + h)
f( x ) A
h f(x + h) – f(x) xx + h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches
Thus,
approaches
the slope of
secant approaches the slope of the tangent to the curve at that point.
slope of
tangent Expressed in limit notation:
Expressed limit The slope of the tangent line to the graph
The
of
of f at the point P(x, f(x)) is given by lim
h→0
if it exists. f ( x + h) − f ( x)
h Average Rates of Change
Average We can see that measuring the slope of the tangent line to
slope of
tangent a graph is mathematically equivalent to finding the
mathematically
rate of change of f at x.
rate The number f(x + h) – f(x) measures the change in y that
The
change
corresponds to a change h in x.
change Then the difference quotient
Then
difference
f ( x + h) − f ( x ) h
measures the average rate of change of y with respect to x
average
over the interval [x, x + h]. In the maglev example, if y measures the position of the
In
maglev
if
position
train at time x, then the quotient gives the average velocity
then
quotient
average
of the train over the time interval [x, x + h]. Average Rates of Change
Average The average rate of change of f over the interval
The average
interval [x, x + h] or slope of the secant line to the graph of
slope
secant
f through the points (x, f(x)) and (x + h, f(x + h)) is f ( x + h) − f ( x )
h Instantaneous Rates of Change
Instantaneous By taking the limit of the difference quotient as h goes to
of zero, evaluating
zero, f ( x + h) − f ( x )
lim
h→0
h
we obtain the rate of change of f at x.
we
rate This is known as the instantaneous rate of change of f at x
This
instantaneous
(as opposed to the average rate of change).
(as
average rate In the maglev example, if y measures the position of a
In
position
train at time x, then the limit gives the velocity of the train
velocity
at time x. Instantaneous Rates of Change
Instantaneous The instantaneous rate of change of f at x or slope
The instantaneous
slope of the tangent line to the graph of f at (x, f(x)) is
of f ( x + h) − f ( x)
lim
h→0
h This limit is called the derivative of f at x .
This
derivative The Derivative of a Function
The The derivative of a function f with respect to x is the
The function f′′ (read “f prime”).
(read f ( x + h) − f ( x)
f ′( x ) = lim
h→0
h The domain of f ′ is the set of all x where the limit exists.
The domain
all
limit Thus, the derivative of function f is a function f ′ that
Thus,
derivative
function gives the slope of the tangent line to the graph of f at any
slope of
point (x, f(x)) and also the rate of change of f at x.
rate The Derivative of a Function
The Four Step Process for Finding f ′(x)
Four 1. Compute f(x + h).
Compute
2. Form the difference f(x + h) – f(x).
Form f ( x + h) − f ( x)
.
3. Form the quotient
3.
h
f ( x + h) − f ( x)
′( x ) = lim
.
4. Compute f
Compute
h→0
h Examples
Examples Find the slope of the tangent line to the graph f(x) = 3x + 5
Find
slope at any point (x, f(x)).
Solution The required slope is given by the derivative of f at x.
The
derivative To find the derivative, we use the fourstep process:
To
fourstep
Step 1. f(x + h) = 3(x + h) + 5 = 3x + 3h + 5.
Step 2. f(x + h) – f(x) = 3x + 3h + 5 – (3x + 5) = 3h.
Step 3. f ( x + h ) − f ( x ) 3h
=
= 3.
h
h Step 4. f ′( x ) = lim f ( x + h ) − f ( x ) = lim 3 = 3.
Step
h→0
h→0
h Examples
Examples Find the slope of the tangent line to the graph f(x) = x2 at
Find
slope any point (x, f(x)).
Solution The required slope is given by the derivative of f at x.
The
derivative To find the derivative, we use the fourstep process:
To
fourstep
Step 1. f(x + h) = (x + h)2 = x2 + 2xh + h2.
Step 2. f(x + h) – f(x) = x2 + 2xh + h2 – x2 = h(2x + h).
Step 3. f ( x + h ) − f ( x ) h(2 x + h )
=
= 2 x + h.
h
h Step 4. f ′( x ) = lim f ( x + h ) − f ( x ) = lim(2 x + h ) = 2 x .
Step
h→0
h→0
h Examples
Examples Find the slope of the tangent line to the graph f(x) = x2 at
Find
slope any point (x, f(x)). The slope of the tangent line is given by f ′(x) = 2x. Now, find and interpret f ′(2).
Now, find
y
Solution
f ′(2) = 2(2) = 4.
5 This means that, at the point (2,
This
at
(2, 4)
4
4)…
4)
3
… the slope of the tangent line
the
to the graph is 4.
2 4 1 1
–2 –1 0 1 2 x Applied Example: Demand for Tires
Applied The management of Titan Tire Company has determined
The that the weekly demand function of their Super Titan tires
weekly
is given by
is
p = f ( x ) = 144 − x 2
where p is measured in dollars and x is measured in
where
dollars
thousands of tires.
thousands Find the average rate of change in the unit price of a tire if
Find
average
unit
the quantity demanded is between 5000 and 6000 tires;
5000
6000
between 5000 and 5100 tires; and between 5000 and 5010
5000
5100
5000
5010
tires.
tires. What is the instantaneous rate of change of the unit price
What
instantaneous
when the quantity demanded is 5000 tires?
5000 Applied Example: Demand for Tires
Applied
Solution The average rate of change of the unit price of a tire if the
The average
quantity demanded is between x and x + h is
f ( x + h ) − f ( x ) [144 − ( x + h )2 ] − (144 − x 2 )
=
h
h
144 − x 2 − 2 xh − h 2 − 144 + x 2
=
h
−2 xh − h 2 h( −2 x − h )
=
=
h
h
= −2 x − h Applied Example: Demand for Tires
Applied
Solution The average rate of change is given by –2x – h.
The average
–2 To find the average rate of change of the unit price of a
To
average
tire when the quantity demanded is between 5000 and
quantity
5000
6000 tires [5, 6], we take x = 5 and h = 1, obtaining
6000
[5,
we
−2(5) − 1 = −11
or –$11 per 1000 tires.
or –$11
1000 Similarly, with x = 5, and h = 0.1, we obtain
Similarly,
and
we
−2(5) − 0.1 = −10.1
or –$10.10 per 1000 tires.
–$10.10
1000 Finally, with x = 5, and h = 0.01, we get
Finally,
and
we
−2(5) − 0.01 = −10.01
or –$10.01 per 1000 tires.
–$10.01
1000 Applied Example: Demand for Tires
Applied
Solution The instantaneous rate of change of the unit price of a tire
The instantaneous
when the quantity demanded is x tires is given by
f ′( x ) = lim
h→0 f ( x + h) − f ( x)
= lim( −2 x − h ) = −2 x
h→0
h In particular, the instantaneous rate of change of the price
the per tire when quantity demanded is 5000 is given by –2(5),
when
5000
–2(5)
or –$10 per tire.
–$10 Differentiability and Continuity
Differentiability Sometimes, one encounters continuous functions that fail
Sometimes,
encounters to be differentiable at certain values in the domain of the
to
domain
function f. For example, consider the continuous function f below:
consider
✦ It fails to be differentiable at x = a , because the graph
It fails
makes an abrupt change (a corner) at that point.
abrupt
(It is not clear what the slope is at that point)
slope is
y a x Differentiability and Continuity
Differentiability Sometimes, one encounters continuous functions that fail
Sometimes,
encounters to be differentiable at certain values in the domain of the
to
domain
function f. For example, consider the continuous function f below:
consider
✦ It also fails to be differentiable at x = b because the
It
fails
slope is not defined at that point.
slope y a b x Applied Example: Wages
Applied Mary works at the B&O department store, where, on a
Mary weekday, she is paid $8 an hour for the first 8 hours and
paid $8
first
$12 an hour of overtime.
$12
overtime The function
if 0 ≤ x ≤ 8
8 x
f ( x) = 12 x − 32 if x > 8 gives Mary’s earnings on a weekday in which she worked
gives
earnings
x hours.
hours Sketch the graph of the function f and explain why it is not
differentiable at x = 8.
differentiable Applied Example: Wages
Applied
Solution
130
110 y
if 0 ≤ x ≤ 8
8 x
f ( x) = 12 x − 32 if x > 8 90
70 (8, 64) 50
30
10
2 4 6 8 10 12 x The graph of f has a corner at x = 8 and so is not
The
corner differentiable at that point.
differentiable 9.4
9.4
Basic Rules of Differentiation
d
( 4 x 5 + 3 x 4 − 8 x 2 + x + 3)
dx
d
d
d
d
d
= 4 ( x 5 ) + 3 ( x 4 ) − 8 ( x 2 ) + ( x ) + ( 3)
dx
dx
dx
dx
dx f ′( x ) = = 4 ( 5x 4 ) + 3 ( 4 x3 ) − 8 ( 2 x ) + 1 + 0
= 20 x 4 + 12 x 3 − 16 x + 1 Four Basic Rules
Four We’ve learned that to find the rule for the derivative f ′of a
We’ve
derivative of function f, we first find the difference quotient
we
difference lim
h→0 f ( x + h) − f ( x )
h But this method is tedious and time consuming, even for
But
and
even relatively simple functions.
relatively This chapter we will develop rules that will simplify the
This
process of finding the derivative of a function.
process Rule 1: Derivative of a Constant
Rule We will use the notation
We d
[ f ( x )]
dx to mean “the derivative of f with respect to x at x.”
“the
with
Rule 1: Derivative of a constant
d
( c) = 0
dx The derivative of a constant function is equal to zero. Rule 1: Derivative of a Constant
Rule We can see geometrically why the derivative of a constant
We
geometrically must be zero.
must The graph of a constant function is a straight line parallel
The
constant
to the x axis.
to
axis Such a line has a slope that is constant with a value of zero.
Such
slope
zero Thus, the derivative of a constant must be zero as well.
y
f( x ) = c x Rule 1: Derivative of a Constant
Rule We can use the definition of the derivative to
We
definition demonstrate this:
demonstrate f ( x + h) − f ( x)
h→0
h
c−c
= lim
h→0 h
= lim 0 f ′( x ) = lim h→0 =0 Rule 2: The Power Rule
Rule Rule 2: The Power Rule If n is any real number, then
If
dn
x ) = nx n −1
(
dx Rule 2: The Power Rule
Rule Lets verify this rule for the special case of n = 2.
Lets If f(x) = x2, then
If f ′( x ) = d2
f ( x + h) − f ( x )
x ) = lim
(
h→0
dx
h ( x + h )2 − x 2
x 2 + 2 xh + h 2 − x 2
= lim
= lim
h→0
h→0
h
h
2 xh + h 2
h(2 x + h )
= lim
= lim
h→0
h→0
h
h
= lim(2 x + h ) = 2 x
h→0 Rule 2: The Power Rule
Rule
Practice Examples: If f(x) = x, then
If If f(x) = x8, then
If If f(x) = x5/2, then
If d
f ′( x ) = ( x ) = 1 ⋅ x1−1 = x 0 = 1
dx
f ′( x ) = d8
( x ) = 8 ⋅ x8−1 = 8 x 7
dx f ′( x ) = d 5/2
5 5/2−1 5 3/2
( x ) = 2⋅x = 2 x
dx Rule 2: The Power Rule
Rule
Practice Examples: Find the derivative of f ( x ) = x
d
f ′( x ) =
dx () d 1/2
x = (x )
dx 1 1/2−1
=x
2
= 1
2x 1 −1/2
=x
2 Rule 2: The Power Rule
Rule
Practice Examples:
1 Find the derivative of f ( x ) =
Find
3
x
f ′( x ) = d 1 d −1/3 3 = dx ( x )
dx x 1
= − x −1/3−1
3
1 −4 / 3
1
=− x
= − 4/3
3
3x Rule 3: Derivative of a Constant Multiple Function
Rule Rule 3: Derivative of a Constant Multiple Function If c is any constant real number, then
If
d
d
cf ( x )] = c [ f ( x )]
[
dx
dx Rule 3: Derivative of a Constant Multiple Function
Rule
Practice Examples: Find the derivative of f ( x ) = 5 x 3
d
f ′( x ) = ( 5 x 3 )
dx
d3
=5 (x )
dx
= 5 ( 3x 2 )
= 15 x 2 Rule 3: Derivative of a Constant Multiple Function
Rule
Practice Examples:
Practice
3 Find the derivative of f ( x ) =
x
f ′( x) = d
( 3x −1/ 2 )
dx 1 −3/ 2 = 3 − x 2 3
= − 3/ 2
2x Rule 4: The Sum Rule
Rule Rule 4: The Sum Rule
d
d
d
[ f ( x ) ± g ( x )] = [ f ( x )] ± [ g ( x )]
dx
dx
dx Rule 4: The Sum Rule
Rule
Practice Examples: Find the derivative of f ( x ) = 4 x 5 + 3x 4 − 8 x 2 + x + 3 d
f ′( x ) = ( 4 x 5 + 3x 4 − 8 x 2 + x + 3)
dx
d5
d4
d2
d
d
= 4 ( x ) + 3 ( x ) − 8 ( x ) + ( x ) + ( 3)
dx
dx
dx
dx
dx
= 4 ( 5x 4 ) + 3 ( 4 x3 ) − 8 ( 2 x ) + 1 + 0
= 20 x 4 + 12 x 3 − 16 x + 1 Rule 4: The Sum Rule
Rule
Practice Examples: Find the derivative of t2 5
g (t ) = + 3
5t d t2 5 d 1 2 g ′(t ) = + 3 = t + 5t −3 dt 5 t dt 5 1d 2
d −3
= ⋅ (t ) +5 (t )
5 dt
dt
1
( 2t ) + 5 ( −3t −4 )
5
2t 15 2t 5 − 75
= − 4=
5t
5t 4
= Applied Example: Conservation of a Species
Applied A group of marine biologists at the Neptune Institute of
group Oceanography recommended that a series of conservation
measures be carried out over the next decade to save a
measures
certain species of whale from extinction.
certain After implementing the conservation measure, the
After
population of this species is expected to be
population
N (t ) = 3t 3 + 2t 2 − 10t + 600 (0 ≤ t ≤ 10) where N(t) denotes the population at the end of year t.
where
population Find the rate of growth of the whale population when
Find
rate
t = 2 and t = 6. How large will the whale population be 8 years after
How
population
implementing the conservation measures?
implementing Applied Example: Conservation of a Species
Applied
Solution The rate of growth of the whale population at any time t is
The rate
whale
given by
given
N ′(t ) = 9t 2 + 4t − 10 In particular, for t = 2, we have
In N ′(2) = 9 ( 2 ) + 4 ( 2 ) − 10 = 34
2 And for t = 6, we have
And N ′(6) = 9 ( 6 ) + 4 ( 6 ) − 10 = 338
2 Thus, the whale population’s rate of growth will be 34
Thus,
rate
34 whales per year after 2 years and 338 per year after 6 years.
338 Applied Example: Conservation of a Species
Applied
Solution The whale population at the end of the eighth year will be
The whale
eighth
N ( 8) = 3 ( 8) + 2 ( 8) − 10 ( 8) + 600
3 2 = 2184 whales 9.5
9.5
The Product and Quotient Rules d
[ f ( x ) g ( x )] = f ( x ) g ′( x ) + g ( x ) f ′( x )
dx
d f ( x ) g ( x ) f ′( x ) − f ( x ) g ′( x )
2 g ( x) =
dx g ( x )]
[ Rule 5: The Product Rule
Rule The derivative of the product of two differentiable
The functions is given by
functions
d
[ f ( x ) g ( x )] = f ( x ) g ′( x ) + g ( x ) f ′( x )
dx Rule 5: The Product Rule
Rule
Practice Examples: Find the derivative of f ( x ) = ( 2 x 2 − 1) ( x 3 + 3)
f ′( x ) = ( 2 x 2 − 1) d3
d
x + 3) + ( x 3 + 3) ( 2 x 2 − 1)
(
dx
dx = ( 2 x 2 − 1) ( 3x 2 ) + ( x 3 + 3) ( 4 x )
= 6 x 4 − 3x 2 + 4 x 4 + 12 x
= x ( 10 x 3 − 3x + 12 ) Rule 5: The Product Rule
Rule
Practice Examples: Find the derivative of f ( x ) = x 3 ( ) x +1 d 1/2
d3
1/2
f ′( x ) = x
( x + 1) + ( x + 1) dx x
dx
3 1 −1/2 = x x + ( x1/2 + 1) 3x 2
2 3 1 5/2
x + 3x 5/2 + 3x 2
2
7
= x 5/2 + 3x 2
2
= Rule 6: The Quotient Rule
Rule The derivative of the quotient of two differentiable
The functions is given by
functions
d f ( x) g ( x) f ′( x) − f ( x) g ′( x) g ( x) =
2
dx g ( x) ] [ ( g ( x ) ≠ 0) Rule 6: The Quotient Rule
Rule
Practice Examples: Find the derivative of x
f ( x) =
2x − 4 d
d
( 2 x − 4) ( x) − x ( 2 x − 4)
dx
dx
f ′( x ) =
2
2 x − 4)
( ( 2 x − 4 ) ( 1) − x ( 2 )
=
2
2 x − 4)
(
= 2x − 4 − 2x ( 2 x − 4) 2 =− 4 ( 2 x − 4) 2 Rule 6: The Quotient Rule
Rule
Practice Examples: Find the derivative of x2 + 1
f ( x) = 2
x −1 d2
d2
2
( x − 1) dx ( x + 1) − ( x + 1) dx ( x − 1)
f ′( x ) =
2
2
( x − 1)
2 (x
=
= 2 − 1) ( 2 x ) − ( x 2 + 1) ( 2 x ) (x 2 − 1) 2 2 x3 − 2 x − 2 x3 − 2 x (x 2 − 1) 2 =− (x 4x
2 − 1) 2 Applied Example: Rate of Change of DVD Sales
Applied The sales ( in millions of dollars) of DVDs of a hit movie
The sales t years from the date of release is given by
5t
S (t ) = 2
t +1 Find the rate at which the sales are changing at time t.
Find
rate
sales How fast are the sales changing at: ✦ The time the DVDs are released (t = 0)?
The
0)
✦ And two years from the date of release (t = 2)?
And
2) Applied Example: Rate of Change of DVD Sales
Applied
Solution The rate of change at which the sales are changing at
The rate
sales
time t is given by
d 5t S ′(t ) = 2 dt t + 1 (t
=
= 2 + 1) ( 5) − ( 5t ) ( 2t ) (t 2 + 1) 5t + 5 − 10t
2 (t 2 + 1) 2 2 2 = 5( 1 − t2 ) (t 2 + 1) 2 Applied Example: Rate of Change of DVD Sales
Applied
Solution The rate of change at which the sales are changing when
The rate
the DVDs are released (t = 0) is
the
1 − ( 0 ) 2 5 1
5 = ( ) =5
′(0) = S
2
2
2
( 1)
( 0 ) + 1 That is, sales are increasing by $5 million per year.
That
increasing
$5 Applied Example: Rate of Change of DVD Sales
Applied
Solution The rate of change two years after the DVDs are
The rate
released (t = 2) is
released
1 − ( 2 ) 2 5 1 − 4
5
) = − 15 = − 3 = −0.6 = (
S ′(2) =
2
2
2
25
5
4 + 1)
(
( 2 ) + 1 That is, sales are decreasing by $600,000 per year.
That
decreasing
$600,000 HigherOrder Derivatives
HigherOrder The derivative f ′ of a function f is also a function.
The
is As such, f ′ may also be differentiated.
As
may Thus, the function f ′ has a derivative f ″ at a point x in the
Thus, domain of f if the limit of the quotient
domain
f ′( x + h ) − f ′( x )
h
exists as h approaches zero.
exists
approaches The function f ″ obtained in this manner is called the
The
second derivative of the function f, just as the derivative f ′
second
just
of f is often called the first derivative of f.
first By the same token, you may consider the third, fourth,
By
third fourth
fifth, etc. derivatives of a function f.
fifth etc. HigherOrder Derivatives
HigherOrder
Practice Examples: Find the third derivative of the function f(x) = x2/3 and
Find
third
determine its domain.
domain
Solution
2
2 1
2 We have f ′( x ) = x −1/3 and f ′′( x ) = − x −4/3 = − x −4/3
We
3
3 3
9 So the required derivative is 24
8 −7/3
8
′′′( x ) = − − x −7/3 =
f
x= 9 3
27
27 x 7/3 The domain of the third derivative is the set of all real
The domain
third
numbers except x = 0.
numbers HigherOrder Derivatives
HigherOrder
Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2
Find
second
(2
Solution Using the general power rule we get the first derivative:
general
first
1/2
1/2
3
2
2
f ′( x ) = ( 2 x + 3) ( 4 x ) = 6 x ( 2 x + 3)
2 HigherOrder Derivatives
HigherOrder
Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2
Find
second
(2
Solution Using the product rule we get the second derivative:
product
second
1/2
1/2
d
d
2
2
f ′′( x ) = 6 x ⋅ ( 2 x + 3) + ( 2 x + 3) ⋅ ( 6 x )
dx
dx
−1/2
1/2
1
2
2
= 6 x ⋅ ( 2 x + 3) ( 4 x ) + ( 2 x + 3) ⋅ 6
2
= 12 x ( 2 x + 3) −1/2 = 6 ( 2 x + 3) 2 x 2 + ( 2 x 2 + 3) 2 2 2 = 6 ( 4 x 2 + 3)
2x2 + 3 −1/2 + 6 ( 2 x + 3)
2 1/2 Applied Example: Acceleration of a Maglev
Applied The distance s (in feet) covered by a maglev moving along
The distance
feet covered a straight track t seconds after starting from rest is given
seconds
by the function
s = 4t2
(0 ≤ t ≤ 10)
(0 What is the maglev’s acceleration after 30 seconds?
What
acceleration
30 Solution The velocity of the maglev t seconds from rest is given by
The velocity
ds d
= ( 4t 2 ) = 8t
dt dt The acceleration of the maglev t seconds from rest is given
The acceleration
by the rate of change of the velocity of t, given by
rate
velocity
v= d
d ds d 2 s d
a = v = = 2 = ( 8t ) = 8
dt
dt dt dt
dt or 8 feet per second per second (ft/sec2).
or 9.6
9.6
The Chain Rule d
h′( x ) =
g ( f ( x ) ) = g ′ [ f ( x )] f ′( x )
dx
dy dy du
=
⋅
dx du dx Deriving Composite Functions
Deriving ( ) Consider the function h ( x ) = x 2 + x + 1 2 To compute h′(x), we can first expand h(x)
To
we
expand h( x ) = ( x + x + 1) = ( x 2 + x + 1) ( x 2 + x + 1)
2 2 = x 4 + 2 x 3 + 3x 2 + 2 x + 1
and then derive the resulting polynomial
derive
h′( x ) = 4 x 3 + 6 x 2 + 6 x + 2 But how should we derive a function like H(x)?
But H ( x ) = ( x + x + 1)
2 100 Deriving Composite Functions
Deriving
Note that H ( x ) = ( x + x + 1)
Note
2 100 is a composite function:
composite H(x) is composed of two simpler functions f ( x) = x2 + x + 1 g ( x ) = x100 and So that H ( x ) = g [ f ( x )] = [ f ( x )] 100 = ( x + x + 1)
2 100 We can use this to find the derivative of H(x). Deriving Composite Functions
Deriving
To find the derivative of the composite function H(x):
To find
composite We let u = f(x) = x2 + x + 1 and y = g(u) = u100.
We Then we find the derivatives of each of these functions
Then
find du
= f ′( x ) = 2 x + 1
dx and dy
= g ′(u ) = 100u 99
du The ratios of these derivatives suggest that
The ratios dy dy du
=
⋅
= 100u 99 ( 2 x + 1)
dx du dx Substituting x2 + x + 1 for u we get
we
99
dy
2
H ′( x ) =
= 100 ( x + x + 1) ( 2 x + 1)
dx Rule 7: The Chain Rule
Rule If h(x) = g[f(x)], then
If h′( x) = d
g ( f ( x) ) = g ′ ( f ( x) ) f ′( x)
dx Equivalently, if we write y = h(x) = g(u),
Equivalently, where u = f(x), then
dy dy du
=
⋅
dx du dx The Chain Rule for Power Functions
The Many composite functions have the special form
Many composite
special h(x) = g[f(x)]
where g is defined by the rule
where
g(x) = xn
so that
so
h(x) = [f(x)]n (n, a real number) In other words, the function h is given by the power of a
In
the function f.
function Examples: h( x ) = ( x + x + 1)
2 100 H ( x) = 1 (5− x ) 33 G( x) = 2 x2 + 3 The General Power Rule
The If the function f is differentiable and
If h(x) = [f(x)]n
then (n, a real number),
real d
n
n −1
h′( x ) = [ f ( x )] = n [ f ( x )] f ′( x )
dx The General Power Rule
The
Practice Examples: Find the derivative of G ( x ) = x 2 + 1
Find
Solution
Solution
1/2 Rewrite as a power function: G ( x ) = ( x 2 + 1)
Rewrite
power Apply the general power rule:
Apply
general
−1/2 d
12
G ′( x ) = ( x + 1)
x 2 + 1)
(
2
dx
−1/2
12
= ( x + 1) ( 2 x )
2
x
=
x2 + 1 The General Power Rule
The
Practice Examples:
5 Find the derivative of f ( x ) = x 2 ( 2 x + 3)
Find
Solution
Solution Apply the product rule and the general power rule:
Apply
product
general
f ′( x ) = x 2 d
5
5d
( 2 x + 3) + ( 2 x + 3) x 2
dx
dx = x ( 5) ( 2 x + 3)
2 4 ( 2 ) + ( 2 x + 3) ( 2 ) x = 10 x 2 ( 2 x + 3) + 2 x ( 2 x + 3)
4 = 2 x ( 2 x + 3) ( 5x + 2 x + 3)
4
= 2 x ( 2 x + 3) ( 7 x + 3)
4 5 5 The General Power Rule
The
Practice Examples: Find the derivative of f ( x ) =
Find ( 4x 1
2 − 7) 2 Solution
Solution
−2 Rewrite as a power function: f ( x ) = ( 4 x 2 − 7 )
Rewrite
power Apply the general power rule:
Apply
general
f ′( x ) = −2 ( 4 x − 7 )
2 =− 16 x ( 4x 2 − 7) 3 −3 ( 8x ) The General Power Rule
The
Practice Examples:
3 2x + 1 Find the derivative of f ( x ) = Find 3x + 2 Solution
Solution Apply the general power rule and the quotient rule:
Apply
general
quotient
2 2x + 1 d 2x + 1 f ′( x ) = 3 3x + 2 dx 3x + 2 2x + 1 = 3 3x + 2 2 ( 3x + 2 ) ( 2 ) − ( 2 x + 1) ( 3) 2
( 3x + 2 ) 6 x + 4 − 6 x − 3 3 ( 2 x + 1) 2 2x + 1 = 3
= 2
4 3x + 2 ( 3x + 2 )
3x + 2 )
( 2 Applied Problem: Arteriosclerosis
Applied Arteriosclerosis begins during childhood when plaque
Arteriosclerosis
plaque forms in the arterial walls, blocking the flow of blood
blocking
through the arteries and leading to heart attacks, stroke
arteries
and gangrene.
and Applied Problem: Arteriosclerosis
Applied Suppose the idealized cross section of the aorta is circular
Suppose
cross with radius a cm and by year t the thickness of the plaque is
cm
thickness
h = g(t) cm
cm
then the area of the opening is given by
then
area
A = π (a – h)2 cm2 Further suppose the radius of an individual’s artery is 1 cm
Further
radius
cm (a = 1) and the thickness of the plaque in year t is given by
and
thickness
is
h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm Applied Problem: Arteriosclerosis
Applied Then we can use these functions for h and A
Then h = g(t) = 1 – 0.01(10,000 – t2)1/2 A = f(h) = π (1 – h)2 to find a function that gives us the rate at which A is
rate
changing with respect to time by applying the chain rule:
changing
time
chain
dA dA dh
=
⋅
= f ′( h ) ⋅ g ′(t )
dt dh dt 1
2 −1/2
= 2π (1 − h )( −1) −0.01 ( 10,000 − t ) ( −2t ) 2 0.01t = −2π (1 − h ) 1/2 ( 10,000 − t 2 ) 0.02π (1 − h )t
=−
10,000 − t 2 Applied Problem: Arteriosclerosis
Applied For example, at age 50 (t = 50),
For
50 h = g (50) = 1 − 0.01(10,000 − 2500)1/2 ≈ 0.134 So that dA
0.02π (1 − 0.134)50
=−
≈ −0.03
dt
10,000 − 2500 That is, the area of the arterial opening is decreasing at the
That
area rate of 0.03 cm2 per year for a typical 50 year old.
rate 0.03 cm
50 9.7
9.7
Differentiation of the Exponential
Differentiation
and Logarithmic Functions
and
y (− 1
2 , e −1/2 ) 1 ( 1
2 , e −1/2 )
2
2 f ( x) = e− x x
–1 1 Rule 8
Rule
Derivative of the Exponential Function The derivative of the exponential function with
The base e is equal to the function itself:
dx
( e ) = ex
dx Examples
Examples Find the derivative of the function f ( x ) = x 2 e x
Find
derivative Solution Using the product rule gives
Using
product
d 2x
d
d
x e ) = x2 ( ex ) + ex ( x2 )
(
dx
dx
dx
= x 2e x + e x (2 x ) f ′( x ) = = xe x ( x + 2) Examples
Examples ( Find the derivative of the function g (t ) = e + 2
Find
derivative Solution Using the general power rule gives
Using
general
1/2 d
3t
g ′(t ) = ( e + 2 )
( et + 2 )
2
dt
1/2 t
3t
= ( e + 2) e
2
1/2
3t t
= e ( e + 2)
2 t ) 3/2 Rule 9
Rule
Chain Rule for Exponential Functions If f(x) is a differentiable function, then
If d f ( x)
e ) = e f ( x ) f ′( x )
(
dx Examples
Examples Find the derivative of the function f ( x ) = e 2 x
Find
derivative Solution
d
f ′( x ) = e
( 2x)
dx
= e 2 x (2)
2x = 2e 2 x Examples
Examples Find the derivative of the function y = e −3 x
Find
derivative Solution
dy
−3 x d
=e
( 3 x )
dx
dx
= e −3 x ( −3)
= −3e −3 x Examples
Examples Find the derivative of the function g (t ) = e 2 t
Find
derivative Solution
g ′(t ) = e 2 t 2 +t d
⋅ ( 2t 2 + t )
dt = (4t + 1)e 2 t 2 +t 2 +t Examples
Examples Find the derivative of the function
Find
derivative y = xe −2 x Solution
dy
d −2 x
−2 x d
= x (e ) +e
( x)
dx
dx
dx
d = x e −2 x ( −2 x ) + e −2 x (1)
dx = xe −2 x ( −2) + e −2 x
= −2 xe −2 x + e −2 x
= e −2 x (1 − 2 x ) Examples
Examples et Find the derivative of the function g (t ) =
Find
derivative
et + e − t
Solution
t
−t d
t
td
e +e ) (e ) −e
et + e − t )
(
(
dt
dt
g ′(t ) =
t
−t 2
(e +e ) (e
=
=
= t + e − t ) et − et ( et − e − t ) (e t +e e2t + 1 − e2t + 1 (e (e t +e 2
t +e ) −t 2 ) −t 2 ) −t 2 Rule 10
Rule
Derivative of the Natural Logarithm The derivative of ln x is
The
ln d
1
ln x =
dx
x ( x ≠ 0) Examples
Examples Find the derivative of the function f ( x ) = x ln x
Find
derivative Solution
d
d
f ′( x ) = x ⋅ (ln x ) + ln x ⋅ ( x )
dx
dx
1
= x ⋅ + ln x ⋅ (1) x
= 1 + ln x Examples
Examples Find the derivative of the function g ( x ) =
Find
derivative Solution
d
d
(ln x ) − ln x ⋅ ( x )
dx
g ′( x ) = dx
x2
1
x ⋅ − ln x ⋅ (1)
x
=
x2
1 − ln x
=
x2
x⋅ ln x
x Rule 11
Rule
Chain Rule for Logarithmic Functions If f(x) is a differentiable function, then
If d
f ′( x )
[ ln f ( x )] =
dx
f ( x) [ f ( x ) > 0] Examples
Examples Find the derivative of the function f ( x ) = ln( x 2 + 1)
Find
derivative Solution
d2
( x + 1)
f ′( x ) = dx 2
x +1
2x
=2
x +1 Examples
Examples Find the derivative of the function y = ln[( x 2 + 1)( x 3 + 2)6 ]
Find
derivative Solution y = ln[( x 2 + 1)( x 3 + 2)6 ]
= ln( x 2 + 1) + ln( x 3 + 2)6
= ln( x 2 + 1) + 6ln( x 3 + 2)
d2
d3
( x + 1)
( x + 2)
dy dx
=
+ 6 dx 3
dx
x2 + 1
x +2
2x
3x 2
=2
+6 3
x +1
x +2
2x
18 x 2
=2
+3
x +1 x + 2 9.8
9.8
Marginal Functions in Economics MC(251) = C(251) – C(250)
(251)
(251)
= [8000 + 200(251) – 0.2(251)2]
– [8000 + 200(250) – 0.2(250)2]
[8000
= 45,599.8 – 45,500
= 99.80 Marginal Analysis
Marginal Marginal analysis is the study of the rate of change of economic quantities.
economic These may have to do with the behavior of costs, revenues,
These
profit, output, demand, etc.
profit, In this section we will discuss the marginal analysis of
In
various functions related to:
various
✦ Cost
✦ Average Cost
✦ Revenue
✦ Profit Applied Example: Rate of Change of Cost Functions
Applied Suppose the total cost in dollars incurred each week by
Suppose
total Polaraire for manufacturing x refrigerators is given by the
total cost function
total
C(x) = 8000 + 200x – 0.2x2
(0 ≤ x ≤ 400)
a. What is the actual cost incurred for manufacturing the
What
251st refrigerator?
251 refrigerator?
b. Find the rate of change of the total cost function with
Find
rate
respect to x when x = 250.
c. Compare the results obtained in parts (a) and (b).
Compare
(a)
(b) Applied Example: Rate of Change of Cost Functions
Applied
Solution
a. The cost incurred in producing the 251st refrigerator is
The
251
C(251) – C(250) = [8000 + 200(251) – 0.2(251)2]
(251)
– [8000 + 200(250) – 0.2(250)2]
[8000
= 45,599.8 – 45,500
= 99.80
or $99.80.
or $99.80 Applied Example: Rate of Change of Cost Functions
Applied
Solution
a. The rate of change of the total cost function
The rate
C(x) = 8000 + 200x – 0.2x2
with respect to x is given by
with
C´(x) = 200 – 0.4x
C´
200
So, when production is 250 refrigerators, the rate of
So,
250
change of the total cost with respect to x is
C´(x) = 200 – 0.4(250)
200
= 100
or $100.
or $100 Applied Example: Rate of Change of Cost Functions
Applied
Solution
a. Comparing the results from (a) and (b) we can see they are
Comparing
(a)
(b)
very similar: $99.80 versus $100.
very
$99.80
$100
✦ This is because (a) measures the average rate of change
This
(a)
average
over the interval [250, 251], while (b) measures the
[250,
while (b)
instantaneous rate of change at exactly x = 250.
instantaneous
✦ The smaller the interval used, the closer the average rate
smaller the interval
the closer
of change becomes to the instantaneous rate of change.
of
becomes
instantaneous Applied Example: Rate of Change of Cost Functions
Applied
Solution The actual cost incurred in producing an additional unit
The
additional
of a good is called the marginal cost.
marginal As we just saw, the marginal cost is approximated by the
As
marginal
approximated
rate of change of the total cost function.
rate
total For this reason, economists define the marginal cost
For
function as the derivative of the total cost function.
derivative
total Applied Example: Marginal Cost Functions
Applied A subsidiary of Elektra Electronics manufactures a
subsidiary portable music player.
portable Management determined that the daily total cost of
Management
daily
producing these players (in dollars) is
C(x) = 0.0001x3 – 0.08x2 + 40x + 5000
where x stands for the number of players produced.
where
number
a. Find the marginal cost function.
b. Find the marginal cost for x = 200, 300, 400, and 600.
Find
c. Interpret your results. Applied Example: Marginal Cost Functions
Applied
Solution
a. If the total cost function is:
If
total
C(x) = 0.0001x3 – 0.08x2 + 40x + 5000
then, its derivative is the marginal cost function:
then, its derivative
marginal
C´(x) = 0.0003x2 – 0.16x + 40 Applied Example: Marginal Cost Functions
Applied
Solution
a. The marginal cost for x = 200, 300, 400, and 600 is:
The marginal
C´(200) = 0.0003(200)2 – 0.16(200) + 40 = 20
C´
C´(300) = 0.0003(300)2 – 0.16(300) + 40 = 19
C´(400) = 0.0003(400)2 – 0.16(400) + 40 = 24
C´(600) = 0.0003(600)2 – 0.16(600) + 40 = 52
or $20/unit, $19/unit, $24/unit, and $52/unit, respectively. Applied Example: Marginal Cost Functions
Applied
Solution
a. From part (b) we learn that at first the marginal cost is
From
(b)
at
marginal
decreasing, but as output increases, the marginal cost
decreasing but as
the marginal
increases as well.
increases
This is a common phenomenon that occurs because of
This
common
several factors, such as excessive costs due to overtime and
several
such
overtime
high maintenance costs for keeping the plant running at
maintenance
such a fast rate.
such Applied Example: Marginal Revenue Functions
Applied Suppose the relationship between the unit price p in
Suppose
price dollars and the quantity demanded x of the Acrosonic
quantity
model F loudspeaker system is given by the equation
model
p = – 0.02x + 400 (0 ≤ x ≤ 20,000) a. Find the revenue function R.
Find
revenue
b. Find the marginal revenue function R′.
Find
marginal
c. Compute R′(2000) and interpret your result.
Compute Applied Example: Marginal Revenue Functions
Applied
Solution
a. The revenue function is given by
The revenue
R(x) = px
= (– 0.02x + 400)x
(–
= – 0.02x2 + 400x
0.02 (0 ≤ x ≤ 20,000) Applied Example: Marginal Revenue Functions
Applied
Solution
a. Given the revenue function
Given
revenue
R(x) = – 0.02x2 + 400x
0.02
We find its derivative to obtain the marginal revenue
We
function:
function
R′(x) = – 0.04x + 400 Applied Example: Marginal Revenue Functions
Applied
Solution
a. When quantity demanded is 2000, the marginal revenue
When quantity
the marginal
will be:
will
R′(2000) = – 0.04(2000) + 400
= 320
Thus, the actual revenue realized from the sale of the
Thus,
actual
2001st loudspeaker system is approximately $320.
loudspeaker Applied Example: Marginal Profit Function
Applied Continuing with the last example, suppose the total cost
Continuing
total cost (in dollars) of producing x units of the Acrosonic model F
loudspeaker system is
loudspeaker
C(x) = 100x + 200,000
100
a.
b.
c. Find the profit function P.
Find
profit
Find the marginal profit function P′.
Find
marginal
Compute P′ (2000) and interpret the result.
Compute (2000) Applied Example: Marginal Profit Function
Applied
Solution
a. From last example we know that the revenue function is
From
revenue
R(x) = – 0.02x2 + 400x
✦ Profit is the difference between total revenue and total
difference
total
cost, so the profit function is
cost
P(x) = R(x) – C(x)
= (– 0.02x2 + 400x) – (100x + 200,000)
= – 0.02x2 + 300x – 200,000 Applied Example: Marginal Profit Function
Applied
Solution
a. Given the profit function
Given
profit
P(x) = – 0.02x2 + 300x – 200,000
we find its derivative to obtain the marginal profit
we
derivative
function:
function
P′(x) = – 0.04x + 300
0.04 Applied Example: Marginal Profit Function
Applied
Solution
a. When producing x = 2000, the marginal profit is
When
the marginal
P′(2000) = – 0.04(2000) + 300
0.04(2000)
= 220
220
Thus, the profit to be made from producing the 2001st
Thus,
profit
producing the 2001
loudspeaker is $220.
loudspeaker $220 End of
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This note was uploaded on 07/08/2011 for the course BUS 205 taught by Professor Michael during the Spring '11 term at University of Hawaii, Manoa.
 Spring '11
 michael
 Derivatives

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