2301107 Calculus I
2. Derivatives (I)
Chapter 2:Derivatives 2
Outline
1. The derivatives
1. Definition
2. Derivative as a function
2. Differentiation Formulas 3. Rates of Change
2.1.1. Definition of Derivatives
● Definition: The derivative of a function f at a number a, denoted by f'(a), is
● If we write x = a + h, then h = x – a, f 'a=lim
h0
f ah−f a h
f 'a=lim
xa
f x−f a x−a
Student note
1. Use the definition of limit to find the derivative of a function, f(x) = x2 – 8x + 9 at the number a.
Chapter 2:Derivatives 5
● The tangent line to y = f(x) at (a, f(a)) is the line through (a, f(a)) whose slope is equal to f'(a).
● Estimate the value of f'(0) for f(x) = 2x.
0
0.1 0.718 0.01 0.696 0.001 0.693 -0.1 0.670 -0.01 0.691 -0.001 0.693 x (2^h-1)/h P
y = f(x)
h
f(a+h) - f(a)
Chapter 2:Derivatives 6
Interpretation of the derivative
● The derivative f'(a) is the instantaneous rate of change of y = f(x) with respect to x when x = a.
Student note
2. The position of a particle is given by s= f(t) = 1/(1 + t)
where t is measured in seconds and s in meters. Find the velocity and the speed after 2 seconds.
Student note
3. A manufacturer produces bolts of fabric with a fixed width. The cost of producing x yards of this fabric is C = f(x) dollars.
– What is the meaning of the derivative f'(x)? What are its units?
– In practical terms, what does it mean to say that f'(1000) = 9?
– Which do you think is greater, f'(50) or f'(500)?
Chapter 2:Derivatives 9
Student note
4. If the tangent line to y = f(x) at (4, 3) passes through the point (0, 2), find f(4) and f'(4).
Chapter 2:Derivatives 10
Student note
5. If f(x) = 3x2 – 5x, find f'(2) and use it to find an equation of the tangent line to y = 3x2 – 5x at the point (2, 2).
Student note
6. Find f'(a) 1. f x=2−x4 x2 2. f t=t4−5 t 3. f t=2 t1
t3 4. f x=
3 x1Derivative as a function
● If we let a vary and rewrite a as x,
● Given any number x for which this limit exists, we assign to x the number f'(x), called this new function, the
derivative of f.
f 'x=lim
h0
f xh−f x h
Chapter 2:Derivatives 13
Student note
7. If f(x) = x3 – x, find a formula for f'(x). Draw f and f'.
Chapter 2:Derivatives 14
Student note
8. If , find the derivative of f. State the domain of f'.
f x=
x−1Other notation
● Some common alternative notations for the derivatives
the symbols D and d/dx are called differentiation operators
● Notation for f'(a)
● Definition: A function f is differentiable at a if f'(a) exists. It is differentiable on an open interval (a, b) if it is differentiable at every number in this interval.
f 'x=y '=dy dx =df
dx =d
dx f =D f x=D1 f x=Dx f x
f 'a=dy dx∣x=a
Student note
9. Where is the function f(x) = | x | differentiable?
Chapter 2:Derivatives 17
Fail to be differentiable
● Theorem: If f is differentiable at a, then f is continuous at a.
● Function failed to be differentiable when
– the graph of a function f has a “corner” or “kink” in it;
– any discontinuity of f
– contain vertical tangent line.
0
kink
0
discontinuity
0
Vertical tangent
line
Chapter 2:Derivatives 18
Student note
10.Use the given graph to estimate the value of each derivative. The sketch the graph of f'.
2.1.2. Differentiation formulas
● Derivative of a constant function:
● Derivative of power function:
● The derivative of a constant times a function is the constant times the derivative of the function.
● The Constant Multiple Rule: If c is a constant and f is a differentiable function, then
d
dx
xn
=n xn−1 ddx
c
=0d
dx
c f x
=c ddx f x
Differentiation formulas
● The Sum Rule: If f and g are both differentiable, then
The derivative of a sum of functions is the sum of the derivatives.
● The Difference Rule: If f and g are both differentiable, then
d
dx
f xgx
=ddx f xd
dx gx
d
dx
f x−gx
=ddx f x−d
dx gx
Chapter 2:Derivatives 21
Differentiation formulas
● The Production Rule: If f and g are both differentiable, then
The derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.
d
dx
f xgx
=f xddx gxgxd
dx f x
Chapter 2:Derivatives 22
Differentiation formulas
● The Quotient Rule: If f and g are both differentiable, then
The derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
d
dx
gf xx
=gxdxd f xg−xf2xdxd gxDifferentiation formulas
● General power function: If n is a positive integer, then
● General version of the Power Rule:
d
dx
x−n
=−n x−n−1d
dx
xr
=r xr−1 where r∈ℝStudent note
11. Differentiate the following functions.
1. f x=186.5 2. f x=
303. f x=5 x−1 4. Fx=−4 x10
5. f x=x23 x−4 6. gx=5 x8−2 x56 7. V r=4
3 r2 8. Rt=5 t−
3 5
9. vt=t2− 1
4 t3 10. Rx=
10x7
Chapter 2:Derivatives 25
Student note
12. Differentiate the following functions.
1. f x=2 x33x4−2 x 2. su=u−2u−3u5−2 u2 3. gx=3 x−1
2 x1 4. f t= 2 t
4t2 5. y= t2
3 t2−2 t1 6. y=
x−1
x17. y= r2
1
r 8. y=c x 1c x 9. f x= x
xc x
10. f x=a xb c xd
Chapter 2:Derivatives 26
Student note
13. Find an equation of the tangent line to the curve at the given point.
1. y=2 x
x1 , 1, 1 2. y=
xx1 , 4, 0.4
3. y=x
x , 1, 2 4. y=12 x2, 1, 9Student note
14. If f(3) = 4, g(3) = 2, f'(3) = -6 and g'(3) = 7, find the following numbers.
1. f g'3 2. fg'3
3.
gf
'3 4.
f −f g
'3Student note
15. If where g(4) = 8 and g'(4) = 7, find f'(4).f x=
x gxChapter 2:Derivatives 29
Student note
16. If h(2) = 4 and h'(2) = -3 find d
dx
hxx
∣x=2Chapter 2:Derivatives 30
Student note
17. Find the points on the curve y = x3 – x2 – x + 1 where the tangent line is horizontal.
Student note
18. Find equations of the tangent lines to the curve
that are parallel to the line x - 2y = 2.
y=x−1 x1
Student note
19. Show that the curve y = 6 x3 + 5 x – 3 has no tangent line with slope 4.
Chapter 2:Derivatives 33
Student note
20. A manufacturer produces bolts of fabric with a fixed width. The quantity q of this fabric that is sold is a function of the selling price p, so we can write q = f(p).
Then the total revenue earned with selling price p is R(p)
= p f(p).
(a) What does it mean to say that f(20) = 10,000 and f'(20) = -350?
(b) Assuming the values in part (a), find R'(20) and interpret your answer.
Chapter 2:Derivatives 34
Rates of change
● The derivative of y = f(x) can be interpreted as the rate of change of y with respect to x.
● If s = f(t) is the position function of a particle that is moving in a straight line, then represents the average velocity over a time period ∆t, and v = represents the instantaneous velocity.
s
t ds
dt
Physics application
21. The position of a particle is given by the equation
s = f(t) = t
3– 6 t
2+ 9t
where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
(b) What is the velocity after 2 s? After 4 s?
(c) When is the particle at rest?
(d) When is the particle moving forward (that is, in the positive direction)?
(e) Draw a diagram to represent the motion of the particle.
(f) Find the total distance traveled by the particle during the first five seconds.
Physics application
● A current exists whenever electric charges move. If ∆Q is the net charge that passes through this surface during a time period ∆t, then the average current during this time interval is defined as
average current =
● If we take the limit of this average current over smaller and smaller time intervals, we get what is called the current I at a given time t:
I=lim
t0
Q
t =dQ dt
Q
t
Chapter 2:Derivatives 37
Chemistry application
● A chemical reaction results in the formation of one or more substance (called products) from one or more starting materials (called reactants). Example
2 H2 + O2→ 2 H2O
indicates that two molecules of hydrogen and one molecule of oxygen form two molecules of water.
A + B → C
where A and B are the reactants and C is the product. The concentration of a reactant A is the number of moles per liter and is denoted by [A]. The concentration varies
during a reaction, so [A], [B] and [C] are all functions of time (t).
Chapter 2:Derivatives 38
Chemistry application
● The average rate of reaction of the product C over a time interval t1 < t < t2 is
But chemists are more interested in the instantaneous rate of reaction, which is obtained by taking the limit of the average rate of reaction as the time interval ∆t
approaches 0
rate of reaction =
[C]
t =[C]t2−[C]t1 t2−t1
lim
t0
[C]
t =d[C] dt
Chemistry application
● Since [A] and [B] each decrease at the same rate that [C]
increases, we have rate of reaction =
More generally, it turns out that for a reaction of the form aA + bB → cC + dD
we have
d[C]
dt =−d[A]
dt =−d[B] dt
−1 a
d[A]
dt =−1 b
d[B] dt =1
c
d[C] dt =1
d
d[D] dt
Chemistry application
● One of the quantities of interest in thermodynamics is compressibility. If a given substance is kept at a constant temperature, then its volume V depends on its pressure P.
We can consider the rate of change of volume with respect to pressure – namely, the derivative . As P increases, V decreases, V decreases, so
dV dV dP dP0.
Chapter 2:Derivatives 41
Chemistry application
● The compressibility is defined by introducing a minus sign and dividing this derivative by the volume V:
isothermal compressibility = =−1 V
dV dt
Chapter 2:Derivatives 42
Chemist application
22. The volume V (in cubic meters) of a sample of air at 25◦C was found to be related to the pressure P (in kilopascals) by
V =
The rate of change of V with respect to P when P = 50 kPa is
The compressibility at that pressure is dV
dP∣P=50=−5.3
P2 ∣P=50 =−0.00212 m3/kPa
=−1 V
dV
dt ∣P=50=0.00212 5.3
50
=0.02m3/kPa/m3 5.3
P
Biology application
● Let n = f(t) be the number of individuals in an animal or plant population at time t. The change in the population size between the times t = t1 and t = t2 is ∆n = f(t2) – f(t1), and so the average rate of growth during the time period t1
< t < t2 is
average rate of growth =
The instantaneous rate of growth is obtained from this average rate of growth by letting the time period ∆t approach 0, growth rate =
n
t= f t2−f t1 t2−t1
lim
t0
n
t =dn dt
Biology application
● We consider the flow of blood through a blood vessel, such as a vein or artery, as a cylindrical tube with radius R and length l.
● Because of friction at the walls of the tube, the velocity v of the blood is greatest along the central axis of the tube and decreases as the distance r. At the wall, v becomes 0.
The relationship between v and r is given by the law of laminar flow discovered by the French physician Jean- Louis-Marie Poiseuille in 1840.
v= P
4l R2−r2
Chapter 2:Derivatives 45
Biology application
● where η is the viscosity of the blood and P is the pressure difference between both ends of the tubes.
● If P and l are constant, then v is a function of r with domain [0, R].
● The average rate of change of the velocity as we move from r = r1 outward to r = r2 is given by
if we let ∆r → 0, then the instantaneous rate of change of velocity with respect to r:
v
r=vr2−vr1 r2−r1
lim
r0
v
r=dv dr
Chapter 2:Derivatives 46
Economics application
● Suppose C(x) is the total cost that a company incurs in producing x units of a certain commodity. The function C is called a cost function.
● If the number of items produced is increased from x1 to x2, the additional cost is ∆C = C(x2) – C(x1), and the average rate of change of the cost is
● The limit ∆x → 0, that is, the instantaneous rate of change of cost with respect to the number of items produced, is called the marginal cost
C
x=Cx2−Cx1
x2−x1 =Cx1x−Cx1
x
lim
x0
C
x =dC dx
Student note
23. The position function of a particle is given by s = t3 – 4.5 t2 – 7t, t > 0
When does the particle reach a velocity of 5 m/s?
Student note
24. A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm./s. Find the rate at which the area within the circle is increasing after
(2.1) 1 second (2.2) 3 seconds (2.3) 5 seconds
What can you conclude?
