This Week
Read Sections 3.2-3.4 in Stewart
Homework #3 due 11:30 Thursday evening
worksheet #4 and midterm review in section on Tuesday
Professor Christopher Hoffman
Math 124
There are currently about 106 million private sector jobs. Over
the past few months this number has increased at a rate of
about 340 thousand per year. Currently the average hourly
wage is $22.67 and this is increasing at a rate of $.32/hour per
year. The average number of hours worked is staying constant
at 1768 hours/year.
1
What is the total annual private sector payroll right now?
2
If these rates of increase continue then how much will the
private sector payroll increase by this time next year?
3
If these rates of increase continue then how much will the
private sector payroll increase after t years?
4
What is the instantaneous rate of increase in the total
private sector payroll.
Professor Christopher Hoffman
Math 124
To find the total payroll we multiply the number of employees
times the average salary times the average number of hours
worked.
Payroll
=
(Number of jobs)(Average hourly wage)
(Hours per year)
=
(106 million)(22.67)(1768)
=
4.249 trillion.
A year from now it will be
Payroll
=
(Number of jobs)(Average hourly wage)
(Hours per year)
=
(106.34 million)(22.99)(1768)
=
4.322 trillion.
Thus payroll will have increased by about 73.8 billion.
Professor Christopher Hoffman
Math 124
Total private sector payroll t years from now will be
1768(106 + .34t)(22.67 + .32t)
in millions of dollars per year. The increase in payroll will be
1768(106 + .34t)(22.67 + .32t) − (1768)(106)(22.67)
= (1768) 22.67(.34t) + (106)(.32t) + (.34t)(.32t) .
This will have occurred over t years so the average rate of
change is
(1768) 22.67(.34) + (106)(.32) + (.34)(.32t)
million dollars per year per year.
Taking the limit as t goes to zero we get the instantaneous rate
of change is
(1768) 22.67(.34) + (106)(.32)
million or 73.6
billion
dollars
per year
per year.
Professor
Christopher
Hoffman
Math 124
Product Rule
Theorem
d
f (t)g(t) = f (t)g 0 (t) + g(t)f 0 (t).
dt
Professor Christopher Hoffman
Math 124
Derivation of the Product Rule
If
u(t) ≈ u + u 0 t
and
v (t) ≈ v + v 0 t
then
d
(uv )(0)
dt
≈
u(t)v (t) − u(0)v (0)
t
0
(u + u t)(v + v 0 t) − uv
lim
t
t→0
0
0
uv t + vu t + u 0 v 0 t 2
lim
t
t→0
lim uv 0 + vu 0 + u 0 v 0 t
≈
uv + vu 0
=
≈
≈
lim
t→0
t→0
0
Professor Christopher Hoffman
Math 124
Pictorially we see the product rule by
Professor Christopher Hoffman
Math 124
Find the derivatives of the following functions
f (x) = xe2x
√
g(t) = t(t + bt )
√
√
h(x) = 3x (a x − bx )(2x + 1/ x).
Professor Christopher Hoffman
Math 124
We break the function f (x) = xe2x up into two parts x and e2x .
Then we take the derivatives of the two parts.
d
(x) = 1
dx
and
d 2x
(e ) = 2e2x .
dx
Then by the product rule
d
d
d
(xe2x ) = x (e2x ) + e2x (x) = 2xe2x + e2x .
dx
dx
dx
Professor Christopher Hoffman
Math 124
√
√
We break the function g(t) = t(t + bt ) up into two parts t
and (t + bt ).
Then we take the derivatives of the two parts.
d √
d
1
( t) = (t 1/2 ) = t −1/2 .
dt
dt
2
d
(t + bt ) = 1 + ln(b)bt .
dt
Then by the product rule
g 0 (t)
=
=
√ d
d√
t (t + bt ) + (t + bt )
t
dt
dt √
1 −1/2
t(1 + ln(b)bt ) + (t + bt )
t
.
2
Professor Christopher Hoffman
Math 124
√
√
h(x) = 3x (a x − bx )(2x + 1/ x).
Here we break the function up into two parts
√
√
3x (a x − bx ) and (2x + 1/ x).
For the derivative of the first part we need to use the product
rule.
d x
(3 ) = ln(3)3x
dx
and
d √
a
(a x − bx ) = x −1/2 − ln(b)bx .
dx
2
Then we get
√
d x √
a −1/2
(3 (a x −bx )) = 3x
x
−ln(b)bx +ln(3)3x (a x +bx ).
dx
2
The derivative of the second part is
√
d
1
(2x + 1/ x) = 2 − x −3/2 .
dx
2
Professor Christopher Hoffman
Math 124
Now we get that
d
(h(x))
dx
=
=
√
√
d
3x (a x − bx ) (2x + 1/ x) +
dx
√ d x √
(2x + 1/ x) (3 (a x − bx ))
dx
√
1 −3/2
x
x
3 (a x − b ) 2 − x
+
2
√
√
a −12
(2x + 1/ x) 3x
x
− ln(b)bx + ln(3)3x (a x + bx )
2
Professor Christopher Hoffman
Math 124
Product Rule for three terms
Using the product rule twice we get
Theorem
d
f (t)g(t)h(t)
dt
=
0
0
f (t)g(t)h (t) + f (t)g (t) + g(t)f (t) h(t)
=
f (t)g(t)h0 (t) + f (t)g 0 (t)h(t) + f 0 (t)g(t)h(t)
0
We can apply the product rule as many times as we need.
Professor Christopher Hoffman
Math 124
Quotient Rule
Theorem
d
dt
f (t)
g(t)
=
Professor Christopher Hoffman
g(t)f 0 (t) − f (t)g 0 (t)
.
(g(t))2
Math 124
Proof of the Quotient Rule
Let
h(t) =
f (t)
g(t)
Then
h(t)g(t)
=
f (t)
0
(h(t)g(t))
=
f 0 (t)
h(t)g 0 (t) + g(t)h0 (t)
=
f 0 (t)
g(t)h0 (t)
=
f 0 (t) − h(t)g 0 (t)
0
h (t)
h0 (t)
Professor Christopher Hoffman
f 0 (t) −
=
=
f (t) 0
g(t) g (t)
g(t)
− f (t)g 0 (t)
(g(t))2
g(t)f 0 (t)
Math 124
Find the derivative of
y=
x +1
x2 − 6
We let
f (x) = x + 1
and
f 0 (x) = 1.
g(x) = x 2 − 6
and
g 0 (x) = 2x.
Then by the quotient rule
h0 (x) =
g(x)f 0 (x) − f (x)g 0 (x)
(x 2 − 6)(1) − (x + 1)(2x)
=
.
g 2 (x)
(x 2 − 6)2
Professor Christopher Hoffman
Math 124
Calculate the derivative of ex /x using both the product rule and
the quotient rule.
For the product rule we let f (x) = ex and g(x) = 1/x. Then
d x
−1 1
xex − ex
(e /x) = f (x)g 0 (x) + g(x)f 0 (x) = ex 2 + ex =
.
dx
x
x
x2
For the quotient rule we let F (x) = ex and G(x) = x. Then
d x
G(x)F 0 (x) − F (x)G0 (x)
xex − ex
(e /x) =
=
.
dx
G2 (X )
x2
Professor Christopher Hoffman
Math 124
Find the first and second derivatives of
h(x) =
x2 + 3
.
4x
We let
f (x) = x 2 + 3
g(x) = 4x
and
and
f 0 (x) = 2x.
g 0 (x) = ln(4)4x .
Then by the quotient rule
h0 (x) =
4x (2x) − (x 2 + 3) ln(4)4x
g(x)f 0 (x) − f (x)g 0 (x)
.
=
(4x )2
g 2 (x)
Professor Christopher Hoffman
Math 124
Simplifying we get
h0 (x) =
(2x) − (x 2 + 3) ln(4)
.
4x
We let
F (x) = 2x − ln(4)(x 2 + 3)
G(x) = 4x
and
and
F 0 (x) = 2 − 2 ln(4)x.
G0 (x) = ln(4)4x .
Then by the quotient rule
h00 (x)
=
=
G(x)F 0 (x) − F (x)G0 (x)
G2 (x)
4x (2 − 2 ln(4)x) − (2x − ln(4)(x 2 + 3)) ln(4)4x
.
(4x )2
Professor Christopher Hoffman
Math 124
Here is a graph of f (x) = sin(x).
1
Where is f (x) increasing? decreasing?
2
Where is f 0 (x) the largest?
3
Sketch the graph of f 0 (x).
Professor Christopher Hoffman
Math 124
Compare the graph of f 0 (x) with cos(x).
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Find the derivative of sin(x).
d
(sin(x))
dx
=
=
=
=
sin(x + h) − sin(x)
h
h→0
sin(x) cos(h) + sin(h) cos(x) − sin(x)
lim
h
h→0
sin(x) cos(h) − sin(x) sin(h) cos(x)
lim
+
h
h
h→0
cos(h) − 1
sin(h)
sin(x) lim
+ cos(x) lim
h
h
h→0
h→0
lim
To find the derivative we must evaluate the two limits
cos(h) − 1
h
h→0
lim
Professor Christopher Hoffman
and
Math 124
sin(h)
.
h
h→0
lim
Here is a graph of the function g(h) =
We can see that
cos(h)−1
.
h
cos(h) − 1
= 0.
h
h→0
lim
Professor Christopher Hoffman
Math 124
Here is a graph of the function g(h) =
We can see that
sin(h)
h .
sin(h)
= 1.
h
h→0
lim
Professor Christopher Hoffman
Math 124
Using the previous computation we get
d
(sin(x))
dx
=
cos(h) − 1
sin(h)
+ cos(x) lim
h
h
h→0
h→0
sin(x)(0) + cos(x)(1)
=
cos(x)
=
sin(x) lim
Professor Christopher Hoffman
Math 124
Similarly we can find the derivative of cos(x).
d
(cos(x))
dx
=
cos(x + h) − cos(x)
h
h→0
cos(x) cos(h) − sin(h) sin(x) − cos(x)
lim
h
h→0
cos(x) cos(h) − cos(x) sin(h) sin(x)
lim
−
h
h
h→0
sin(h)
cos(h) − 1
− sin(x) lim
cos(x) lim
h
h
h→0
h→0
cos(x)(0) − sin(x)(1)
=
− sin(x).
=
=
=
=
lim
Professor Christopher Hoffman
Math 124
Now we use the derivative of tan(x).
d
d
sin(x)
(tan(x)) =
dx
dx cos(x)
cos(x)(sin(x))0 − sin(x)(cos(x))0
=
cos2 (x)
cos(x) cos(x) − sin(x)(− sin(x))
=
cos2 (x)
=
=
=
cos2 (x) + sin2 (x)
cos2 (x)
1
cos2 (x)
sec2 (x)
We use the quotient rule and the equality
cos2 (x) + sin2 (x) = 1.
Professor Christopher Hoffman
Math 124
Similarly we get
(sin(x))0
=
cos(x)
0
(cos(x))
=
− sin(x)
0
(tan(x))
=
sec2 (x)
(sec(x))0
=
sec(x) tan(x)
0
(csc(x))
=
− csc(x) cot(x)
0
=
− csc2 (x)
(cot(x))
Professor Christopher Hoffman
Math 124
Find the 25th derivative of f (x) = cos(x). We have that
f 0 (x) = − sin(x),
f 00 (x) = (− sin(x))0 = − cos(x)
f (3) (x) = (− cos(x))0 = sin(x)
f (4) (x) = (sin(x))0 = cos(x) = f (x)
f (5) (x) = (cos(x))0 = − sin(x) = f 0 (x)
We can see that
f (x) = f (4) (x) = f (8) (x) = · · · = f (20) (x) = f (24) (x) = cos(x)
and
f (25) (x) = (f (24) )0 = (cos(x))0 = − sin(x).
Professor Christopher Hoffman
Math 124
Find the derivative of g(x) = cos2 (x).
We write g(x) = cos(x) cos(x) and use the product rule so
g 0 (x) = cos(x)(cos(x))0 + cos(x)(cos(x))0 = 2 cos(x)(− sin(x)).
Notice that
0
(cos(x))2 = 2 cos(x)(− sin(x)) = 2 cos(x)(cos(x))0 .
Professor Christopher Hoffman
Math 124
In the previous example we can replace cos(x) by a general
function f (x).
Find the derivative of g(x) = (f (x))2 .
We use the product rule g(x) = f (x)f (x) and
g 0 (x) = f (x)f 0 (x) + f (x)f 0 (x) = 2f (x)f 0 (x).
Next lecture we will see that we can get the same result with
the chain rule.
Professor Christopher Hoffman
Math 124
Use the equation
cos(a + b) = cos(a) cos(b) − sin(a) sin(b).
to find the derivative of cos(2x).
(cos(2x))0
=
(cos(x) cos(x) − sin(x) sin(x))0
We use the product rule
(cos(x) cos(x))0
=
cos(x)(− sin(x)) + (− sin(x)) cos(x)
=
− 2 sin(x) cos(x)
and
(sin(x) sin(x))0
=
sin(x)(cos(x)) + (sin(x)) cos(x)
=
2 sin(x) cos(x)
Professor Christopher Hoffman
Math 124
Putting this together we get
(cos(2x))0
=
(cos(x) cos(x) − sin(x) sin(x))0
=
− 2 sin(x) cos(x) − 2 sin(x) cos(x)
=
− 4 sin(x) cos(x)
=
− 2 sin(2x)
In the last line we used the equality
sin(2x) = 2 sin(x) cos(x).
We will see how to calculate this derivative with the chain rule.
Professor Christopher Hoffman
Math 124
Find the derivative of
f (x) = (x 2 + 3x) cos(x)
We write g(x) = x 2 + 3x and h(x) = cos(x).
Then g 0 (x) = 2x + 3 and h0 (x) = − sin(x).
f 0 (x) = g(x)h0 (x)+h(x)g 0 (x) = (x 2 +3)(− sin(x))+(2x+3) cos(x).
Professor Christopher Hoffman
Math 124
Find the derivative of
We write f (x) =
√
√
x tan(x)
.
11 − 4x
x tan(x) and g(x) = 11 − 4x.
Then by the product rule
f 0 (x) =
√
1
x sec2 (x) + x −1/2 tan(x).
2
and g 0 (x) = −4.
√
x tan x
11 − 4x
g(x)f 0 (x) − f (x)g 0 (x)
=
g 2 (x)
√
√
(11 − 4x)( x sec2 (x) + tan(x)( 12 x −1/2 )) − ( x tan(x))(−4)
=
.
(11 − 4x)2
d
dx
Professor Christopher Hoffman
Math 124
Find the derivative of
G(x) =
(sec x)2x
= sec(x)2x x −3.1 .
x 3.1
First we rewrite the function.
We write f (x) = sec(x), g(x) = 2x and h(x) = x −3.1 .
Then by the product rule
f 0 (x) = sec(x) tan(x),g 0 (x) = ln(2)2x and h0 (x) = −3.1x −4.1 .
Applying the product rule for three terms we get
(f (x)g(x)h(x))0
=
f 0 (x)g(x)h(x) + f (x)g 0 (x)h(x) + f (x)g(x)h0 (x)
=
sec(x) tan(x)2x x −3.1 + sec(x) ln(2)2x x −3.1
+ sec(x)2x (−3.1)x −4.1
Professor Christopher Hoffman
Math 124
In the last problem we could have used the quotient rule and
found the derivative to be
g 0 (x) =
x 3.1 (2x sec(x) tan(x) + sec(x) ln(2)2x ) − (sec x)2x (3.1x 2.1 )
.
x 6.2
A little algebra shows that these are equivalent.
Professor Christopher Hoffman
Math 124
The longest day of the year is June 21 when Seattle gets 16
hours of daylight. The shortest day of the year is December 21
when Seattle gets 8.42 hours. The number of hours of sunlight
varies sinusoidally. Let t be the number of days since Jan 1.
Find a function S(t) for the number of hours of daylight on
day t.
How much shorter will tomorrow be than today?
Use the definition of the derivative to estimate S 0 (292).
Professor Christopher Hoffman
Math 124
We can calculate that June 21 corresponds with t = 171,
October 20 is t = 292 and December 21 is t = 355.
Since S(t) varies sinusoidally we get that
2π
t + C + D.
S(t) = A sin
B
A=
D=
max − min
16 − 8.42
7.58
=
=
= 3.79.
2
2
2
max + min
16 + 8.42
24.42
=
=
= 12.21.
2
2
2
B = period = 365.
Professor Christopher Hoffman
Math 124
To find C we use that S(171) = 16
16
=
S(171)
π/2 −
16
=
3.79
=
1
=
π/2
=
2π
171
365
− 1.373
2π
3.79 sin
171 + C + 12.21
365
2π
3.79 sin
171 + C
365
2π
sin
171 + C
365
2π
171 + C
365
=
C
≈
C.
Thus
S(t) ≈ 3.79 sin
2π
t − 1.373 + 12.21.
365
Professor Christopher Hoffman
Math 124
Today is t = 292 so we are asking for
S(292) − S(293) = −.0566.
This is 3 minutes and 39 seconds less daylight.
S(292.01) − S(292)
≈ −.0568
.01
which is 3 minutes and 41 seconds.
S 0 (292) ≈
Professor Christopher Hoffman
Math 124
Consider the function y = cos(x 2 + 3).
Let f (x) = cos(x) and g(x) = x 2 + 3.
x
2.2
2.1
2.001
1.9
1.99
1.997
f 0 (g(2))g 0 (2)(x − 2)
-.526
-0.324
-0.002634
0.1932
0.02560678
0.00782
Professor Christopher Hoffman
f (g(x)) − f (g(2))
-.734
-0.2628
-0.002628
0.2627
0.0262
0.00788
Math 124
Theorem (Chain Rule)
d
(f (g(x))) = f 0 (g(x))g 0 (x)
dx
This is the most important tool that we have to calculate
derivatives.
Professor Christopher Hoffman
Math 124
Write the following function as the composition f (g(x)) of two
functions. Then find its derivative.
y = e5x−3
f (x) = ex
f 0 (x) = ex
and
and
g(x) = 5x − 3
f 0 (g(x)) = e5x−3
and
Then by the chain rule
y 0 = f 0 (g(x))g 0 (x) = e5x−3 (5).
Professor Christopher Hoffman
Math 124
g 0 (x) = 5
Write the following function as the composition f (g(x)) of two
functions. Then find its derivative.
y = cos(7 − ax)
f (x) = cos(x)
f 0 (x) = − sin(x)
and
and
g(x) = 7 − ax.
f 0 (g(x)) = − sin(7−ax)
and
Then by the chain rule
y 0 = f 0 (g(x))g 0 (x) = − sin(7 − ax)(−a).
Professor Christopher Hoffman
Math 124
g 0 (x) = −a
Write the following function as the composition f (g(x)) of two
functions. Then find its derivative.
z = (2 − 3 cos(x))3
f (x) = x 3
f 0 (x) = 3x 2 ,
and
g(x) = 2 − 3 cos(x)
f 0 (g(x)) = 3(2−3 cos(x))2
and
g 0 (x) = 3 sin(x)
Then by the chain rule
z 0 = f 0 (g(x))g 0 (x) = 3(2 − 3 cos(x))2 (3 sin(x)).
Professor Christopher Hoffman
Math 124
Write the following function as the composition f (g(x)) of two
functions. Then find its derivative.
y = tan(3 − 5 cos(πx))
f (x) = tan(x)
f 0 (x) = sec2 (x)
and
and
g(x) = 3 − 5 cos(πx)
f 0 (g(x)) = sec2 (3 − 5 cos(πx)).
To find g 0 (x) we need to use the chain rule.
g 0 (x) = −5(cos(πx))0 = −5(− sin(πx))(πx)0 = 5 sin(πx)π.
Then by the chain rule
y 0 = f 0 (g(x))g 0 (x) = sec2 (3 − 5 cos(πx))(5 sin(πx))π
Professor Christopher Hoffman
Math 124
Find the derivative of y = (x − x 2 )5 .
f (x) = x 5
f 0 (x) = 5x 4
and
and
g(x) = x − x 2
f 0 (g(x) = 5(x −x 2 )4
and
g(x) = 1−2x.
Then by the chain rule
y 0 = f 0 (g(x))g 0 (x) = 5(x − x 2 )4 (1 − 2x).
Professor Christopher Hoffman
Math 124
Find the derivative of y = (g(x))2 .
We have f (x) = x 2 , f 0 (x) = 2x and f 0 (g(x)) = 2g(x).
Then
y 0 = f 0 (g(x))g 0 (x) = 2g(x)g 0 (x).
Find the derivative of y = (g(x))n .
We have f (x) = x n , f 0 (x) = nx n−1 and f 0 (g(x)) = n(g(x))n−1 .
Then
y 0 = f 0 (g(x))g 0 (x) = n(g(x))n−1 g 0 (x).
Professor Christopher Hoffman
Math 124
Find the derivative of f (x) =
e5−2x
.
cos(x 2 −5x)
We will use the quotient rule.
0
0
cos(x 2 − 5x) e5−2x − e5−2x cos(x 2 − 5x)
f (x) =
.
cos2 (x 2 − 5x)
0
e5−2x
cos(x 2 − 5x)
f 0 (x) =
0
0
=
= e5−2x (5 − 2x)0 = e5−2x (−2)
− sin(x 2 − 5x) (x 2 −5x)0 = (2x−5)(− sin(x 2 −5x
cos(x 2 − 5x)(e5−2x (−2)) − e5−2x (2x − 5)(− sin(x 2 − 5x))
.
cos2 (x 2 − 5x)
Professor Christopher Hoffman
Math 124
√
Find the derivative of f (x) = k sin(x) 3 + ex .
We will use the product rule.
√
0 0 √
f 0 (x) = k sin(x)
3 + ex + k sin(x)
3 + ex .
√
k sin(x)
0
0
= ln(k )k sin(x) (sin(x))0 = ln(k )k sin(x) (cos(x))
1
1
(3 + ex )−1/2 (3 + ex )0 = (3 + ex )−1/2 (ex )
2
2
√
0
sin(x) 1
x −1/2 x
f (x) = k
(3 + e )
e + ln(k )k sin(x) cos(x)
3 + ex .
2
3+
ex
=
Professor Christopher Hoffman
Math 124
Find the derivative of y = ecx using the chain rule.
We write f (x) = ex and g(x) = cx.
Then f 0 (x) = ex and f 0 (g(x)) = ecx and g 0 (x) = c.
Then by the chain rule
y 0 = f 0 (g(x))g 0 (x) = ecx c.
Professor Christopher Hoffman
Math 124