Chapter 2
Differentiation: Basic Concepts
In this Chapter, we will encounter some
important concepts.
 The Derivative
 Product and Quotient Rules, Higher-
Order Derivatives
 The Chain Rule
 Marginal Analysis, Implicit
Differentiation.
1
Review: Techniques of Differentiation
1. Constant Rule:
derivative of a constant is zero
d
c   0
dx
2. Power Rule: for any real n,
d
[ x n ]  nx n 1
dx
3. Constant Multiple Rule: for constant c and a
differentiable function f(x), d cf ( x )  c d  f ( x)
dx
dx
4. Sum Rule: when f(x) and g(x) are both differentiable,
d
d
d
[ f ( x)  g ( x)] 
[ f ( x)] 
[ g ( x)]
dx
dx
dx
Exercise
3x  1
1  4x
1  5x
y
3  2x
y
2
3
Solution
Exercise
Find the equation of the line that is tangent to the
3
2
graph of the functiony   x  5x  3x 1 at
the point (-1,-8).
Suggested Solution:
y  mx  b
t
t
Suppose the equation of the tangent line is
dy
y   x 3  5 x 2  3x  1
m= dx at the point (-1,-8).
dy
m
 3x 2  10x  3
dx
So at the point (-1,-8), m  3* (1)2  10* (1)  3  10
Hence, yt  10xt  b , and therefore -8=(-10)(-1)+b, gives b=2.
The equation of the tangent line to this function at point (-1,-8)
is
y  10x  2
t
t
Review:
1. Relative rate of change:
Relativerateof
changeof Q(x)

 Q( x) dQ / dx
  Q( x)  Q

2. Percentage rate of change:
Percentagerate  100Q( x)
of changeof Q( x)  Q( x) %


Exercise
1.Find the relative rate of change off ( x)  2 x3  5x 2  4
with respect to x for the value x=1.
2.The gross annual earnings of a certain company were
A(t )  0.1t 2  10t  20 thousand dollars t years after its
formation in 2004.
(1)At what rate were the gross annual earnings of the
company growing with respect to time in 2008?
(2)At what percentage rate were the gross annual earnings of
the company growing with respect to time in 2008?
1.Suggested Solution:
3
2
f
(
x
)

2
x

5
x
4
The function is
Its rate of change is f ' ( x)  6 x 2  10x
2
f
'
(
x
)
6
x
 10x
Its relative rate of change is
 3
f ( x) 2 x  5 x 2  4
When x=1, f ' (1)  6 *1  10*1   4  4
f (1) 2 *1  5 * 2  4
1
2.Suggested Solution:
(1)
A(t )  0.1t 2  10t  20
A' (t )  0.2t  10 , which is the rate that company’s gross
annual earning changes (G.A.E) with respect to t years after
2004.
In 2008, t=4, the G.A.E will change at A’(4)=0.2*4+10=10.8
thousand dollars per year.
(2)The percentage rate of that of the company’s G.A.E changes is
A' (t )
expressed as
*100%. So in year 2008, this value is
A(t )
A' (4)
10.8
*100% 
*100%  17.53%
2
A(4)
0.1* 4  10* 4  20
This means the company’s G.A.E increases 17.53% per year
in the year 2008.
Review:
Product Rules: For function y=f(x) and z=g(x). If they are both
differentiable at x, then the derivative of their product is
[ f ( x) g ( x)]' f ' ( x) g ( x)  f ( x) g ' ( x)
d
dy
dz
( y * z)  * z  y *
dx
dx
dx
Quotient Rules: For function y=f(x) and z=g(x). If they are both
differentiable at x, then the derivative of their quotient is
f ( x)
f ' ( x) g ( x)  f ( x) g ' ( x)
[
]' 
g ( x)
[ g ( x)]2
dy
dz
*z  * y
d y
dx
( )  dx
iff z=g(x) ≠ 0
dx z
z2
Review:
Second Order Derivative: is the derivative of the first derivative.
Notation as follow:
2
df
f ' ( x) 
dx
Example:
Function
First Order Derivative
Second Order Derivative
d f
f ' ' ( x)  2
dx
S(t)
S’(t)=V(t)
S”(t)=V’(t)=A(t)
Position
Velocity
Acceleration
Exercise:
An efficiency study of the morning shift at a certain
factory indicates that an average worker arriving on
3
2
the job at 8:00am will have produced
Q(t )  t  8t  15t
units t hours later.
(a)Compute the worker’s rate of production.
(b)At what rate is the worker’s rate of
production changing with respect to time at
9:00am?
Suggested Solution
(a)Worker’s rate of production is
2
P(t )  Q' (t )  3t  16t  15 units per hour
(b) The rate that worker’s rate of production changes is
P' (t )  Q" (t )  6t  16 units per hour per hour
So at 9:00am, t=1
P' (1)  Q" (1)  6  16  10 units per hour per hour
Review:
Chain Rule: For functions y=f(u), which is differentiable at u,
and z=g(x), which is differentiable at x.
The composition function y=f(z)=f(g(x)) is differentiable at x,
which is
dy dy dz

dx dz dx
[ f ( g ( x))]'  f ' ( g ( x)) * g ' ( x)
!!! Pay Attention to the phrase “WITH RESPECT TO”
Exercise
Differentiate
3x  1
f ( x) 
2x 1
Suggested Solution
3x  1
3x  1
 u , where u 
2x 1
2x 1
dy
1
1
then


du 2 u
3x  1
2
2 x-1
du 3(2 x  1)  2(3x  1)
5
and



dx
(2 x  1) 2
(2 x  1) 2
then by chain rule,
f ( x)  y 
dy dy du
5
f ' ( x) 

  (3x  1) 1/ 2 (2 x  1) 3 / 2
dx du dx
2
Review:
Marginal Cost: If C(x) is the total cost of producing x
units of a commodity.
Then the marginal cost of producing x0 units is the
derivative C( x0 ),
which approximates the additional cost C( x0  1)  C( x0 )
incurred when the level of production is increased by one
unit, from x0 to x0  1
Marginal Revenue :
Marginal Profit :
R' ( x0 )
P' ( x0 )  R' ( x0 )  C' ( x0 )
Review:
Approximation by Increment: If y=f(x) is differentiable
at
x  xo , and △x is a small change in x, then
f ( x0  x)  f ( x0 )  f ' ( x0 )x

f  f ( x0  x)  f ( x0 )  f ' ( x0 )x
Approximation Percentage of change: if △x is a small
change in x, then
f
Percentagechangein f  100
f ( x)
f ' ( x)x
 100
f ( x)
Exercise
Use Incremental Approximation to approximate
the value of 1.00380. Remember f(x+x) – f(x)
= f  f’(x)x
f ' ' (0) 2
f ( x)  f (0)  f ' (0) x 
x
2
Suggested solution
•
•
•
•
•
•
•
Let f(x) = x80, x=1, x = 0.003
f’(x) = 80x79.
When x = 1, f’(x) = 80
f  80x = 0.24
f(x+x) – f(x) = 0.24.
1.00380
=1+0.24=1.24
The ratio of Errors (output to input)
• Suppose x is the correct (or precise) input
• And x+x is the incorrect (or approximate)
input.
• Then x is called the error of the input.
• and y = f(x +x) - f(x) is called the error of
the output
• If x is small, then y’  y/ x.
• Or y = y’  x.
21
Example 23
During a medical procedure, the size of a roughly spherical
tumor is estimated by measuring its diameter and using the
formula V  4 R 3 to compute its volume. If the diameter is
3
measured as 2.5 cm with a maximum error of 2%, how
accurate is the volume measurement?
Solution:
A sphere of radius R and diameter x=2R has volume
4
4
x 3
1
1
3
3
V  R   ( )  x   ( 2.5) 3  8.181 cm 3
3
3
2
6
6
The error made in computing this volume using the diameter 2.5， while the actual
diameter is 2.5+△x
， is
V  V (2.5  x)  V (2.5)  V (2.5)x
22
Exercise
1/ 2
Q
(
K
)

600
K
At a certain factory, the daily output is
units,
where K denotes the capital investment measured in
units of $1,000. The current capital investment is
$900,000. Estimate the effect that an additional capital
investment of $800 will have on the daily output.
Suggested Solution
The current capital investment K₀=900 thousand dollars.
The increase in capital investment △K=0.8 thousand dollars.
To estimate the effect of this △K on the daily output:
Q  Q ' ( K )K  300K-1/2K
for K  900, Δ
 0.8
Q  Q ' (900)0.8  300(900) 1/ 2 * 0.8  8 units
This means an additional capital investment of 800 dollars
would increase the daily output by 8 units.
Review:
Differentials:
• Differentials of x is dx=△x (small increment)
• If y=f(x) is a differentiable function of x, then the differential of y
is  f ' ( x)dx
dy
Review:
Implicit Differentiation:
• Explicit Form: a function that can be written as y=f(x)
“y is solved, and given by an equation of x”
• Implicit Form: cannot express y as a equation of x.
• How to differentiate an implicit equation?
(1) Differentiate both side with respect to x
 you now have a function containing x and y and y’.
(2) Express y’ in terms of x and y.
Exercise:
2
2
y

2
xy
 3x 1 implicitly with respect to x.
Differentiate
Exercise:
2
2
Differentiate y  2 xy  3x 1 implicitly with respect to x.
Suggested Solution:
Differentiate on both side, we have:
2 yy'2 y 2  4 xyy'  3
By rearranging theabove equation:
y ' (2 y  4 xy)  3  2 y 2
3  2 y2
y' 
2 y  4 xy
Example
Suppose the output at a certain factory is Q  2x3  x2 y  y3
units, where x is the number of hours of skilled labor used
and y is the number of hours of unskilled labor. The
current labor force consists of 30 hours of skill labor and
20 hours of unskilled labor.
Question: Use calculus to estimate the change in unskilled
labor y that should be made to offset a 1-hour increase in
skilled labor x so that output will be maintained at its
current level.
28
Solution:
Example
The manager of a company determines that when
q hundred units of a particular commodity are
produced, the cost of production is C thousand
dollars, where C 2  3q3  4275. When 1500 units
are being produced, the level of production is
increasing at the rate of 20 units per week.
What is the total cost at this time and at what rate
is it changing?
30
Exercise
When the price of a certain commodity is p dollars per
unit, the manufacturer is willing to supply x thousand
units, where
x 2  2x
p  p 2  31
How fast is the supply changing when the price is $9 per
unit and is increasing at the rate of 20 cents per week?
31
Exercise
A lake is polluted by waste from a plant located on its
shore. Ecologists determine that when the level of
pollutant is x parts per million (ppm), there will be F fish
of a certain species in the lake, where
32000
F 
3 x
When there are 4000 fish left in the lake, the pollution is
increasing at the rate of 1.4ppm/year. At what rate is the
fish population changing at this time?
32