Transcript Section 2.2 Basic Differentiation Rules & Rates of Change

Unit 2 – Differentiation
Objectives:
1.
2.
3.
4.
5.
6.
Find the derivative of a function using the Constant Rule.
Find the derivative of a function using the Power Rule.
Find the derivative of a function using the Constant Multiple Rule.
Find the derivative of a function using the Sum and Difference Rules.
Find the derivatives of the sine function and of the cosine function.
Use derivatives to find rates of change.
» The derivative of a constant function is 0. That
𝑑
is, if 𝑐 is a real number, then
𝑐 = 0.
𝑑𝑥
» Find the derivatives of the following functions using
the Constant Rule.
Function:
Derivative:
A. 𝑦 = 4

B. 𝑓(𝑥) = 0
 𝑓′(𝑥) = 0
C. 𝑠(𝑡) = −8
 𝑠′(𝑡) = 0
D. 𝑦 = 𝑘𝜋𝑒, 𝑘 is constant
 𝑦′ = 0
𝑑𝑦
𝑑𝑥
=0
If 𝑛 is a rational number, then the function 𝑓(𝑥) =
𝑑
𝑛
𝑥 is differentiable and
𝑥 𝑛 = 𝑛𝑥 𝑛−1 .
𝑑𝑥
For 𝑓 to be differentiable at 𝑥 = 0, 𝑛 must be a
number such that 𝑥 𝑛−1 is defined on an interval
containing 0.
» Find the derivatives of the following functions
using the Power Rule.
Derivative:
Function:
A. 𝑦 = 𝑥
5
B. 𝑓(𝑥) = 𝑥
C. 𝑔 𝑥 =
1
𝑥3

𝑑𝑦
𝑑𝑥

𝑓′

𝑔′
= 5𝑥 4
1
2
𝑥 =
𝑑
𝑑𝑥
𝑥
𝑥 =
𝑑
𝑑𝑥
𝑥 −3
=
1 −1
𝑥 2
2
=
=
1
1
2𝑥 2
−3𝑥 −4
=
3
− 4
𝑥
» Find the slope of the graph of 𝑓(𝑥) = 𝑥 3 when:
A. 𝑥 = −1
Recall that the derivative of a
function gives a slope function.
B. 𝑥 = 0
𝑓 ′ 𝑥 = 3𝑥 2
C. 𝑥 = 2
Now, to find the slopes at those
x-values, just substitute them in.
» Find the slope of the graph of 𝑓(𝑥) = 𝑥 3 when:
𝑓 ′ 𝑥 = 3𝑥 2
y
9
A. 𝑥 = −1
𝑓′
8
7
−1 = 3 −1
2
6
=3
5
4
B. 𝑥 = 0
𝑓′ 0 = 3 0
3
2
1
2
=0
–3
–2
–1
–1
1
2
–2
C. 𝑥 = 2
𝑓′ 2 = 3 2
–3
–4
–5
2
= 12
–6
–7
–8
–9
𝑦 = 𝑥3
3
x
» Find an equation of the tangent line to the
graph of 𝑓(𝑥) = 𝑥 4 when 𝑥 = 2.
Again the derivative is needed to find the slope of the tangent line.
Additionally, a point on the line is needed to write the equation.
Point on the tangent line: 2, 𝑓(2) = (2,16)
Derivative of 𝑓 𝑥 : 𝑓 ′ 𝑥 = 4𝑥 3
Slope of the graph at 2,16 : 𝑚 = 𝑓 ′ 2 = 4 2
3
Equation of tangent line using point−slope form:
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − 16 = 32 𝑥 − 2
𝑦 − 16 = 32𝑥 − 64
𝑦 = 32𝑥 − 48
= 32
» If 𝑓 is a differentiable function and 𝑐 is a real
number, then 𝑐𝑓 is also differentiable and
𝑑
𝑐𝑓(𝑥) = 𝑐𝑓 ′ 𝑥 .
𝑑𝑥
» Find the following derivatives using the
Constant Multiple Rule.
1
− 𝑥
2
A. 𝑓 𝑥 =
𝑓′
𝑥 =
1 𝑑
−
2 𝑑𝑥
B. 𝑠 𝑡 =
𝑠′ 𝑡 =
𝑥1
=
1
−
2
1𝑥 0
2𝑡 4
3
2 𝑑
3 𝑑𝑥
𝑡4 =
2
3
3
8𝑡
4𝑡 3 =
3
1
=−
2
1 =
1
−
2
» Find the following derivatives using the
Constant Multiple Rule.
C. 𝑦 =
𝑑𝑦 1 𝑑
=
𝑑𝑥 4 𝑑𝑥
D. 𝑦 =
3
𝑥
4
3
1 𝑑 1
1 1 −2
𝑥 =
𝑥3 = ∙ 𝑥 3 =
4 𝑑𝑥
4 3
1
2
12𝑥 3
=
5
𝑥
𝑑 1
𝑑 −1
5
−2
𝑦′ = 5
=5
𝑥
= 5 ∙ −1 𝑥 = − 2
𝑑𝑥 𝑥
𝑑𝑥
𝑥
1
3
12 𝑥 2
» Find the following derivatives.
A. 𝑓(𝑥) =
𝑓′
1
4 𝑥3
1 𝑑
𝑥 =
4 𝑑𝑥
B. 𝑦 = 7𝑥
1
𝑥3
1 𝑑 −3
1
3 −5
3
=
𝑥 2 = ∙− 𝑥 2 =− 5
4 𝑑𝑥
4
2
8𝑥 2
6
5
1
6
𝑑𝑦
𝑑
6 1 42𝑥 5
=7
𝑥5 = 7 ∙ 𝑥5 =
𝑑𝑥
𝑑𝑥
5
5
» The sum (or difference) of two (or more)
differentiable functions 𝑓 and 𝑔 is itself
differentiable. Moreover, the derivative of 𝑓 +
𝑔 (or 𝑓 − 𝑔) is the sum (or difference) of the
derivatives of 𝑓 and 𝑔.
𝑑
𝑓 𝑥 ± 𝑔(𝑥) = 𝑓′(𝑥) ± 𝑔′(𝑥)
𝑑𝑥
» Find the following derivatives using the Sum
and Difference Rules.
A. 𝑓 𝑥 = 5𝑥 4 + 2𝑥 2 − 5
𝑓 ′ 𝑥 = 5 4𝑥 3 + 2 2𝑥 = 20𝑥 3 + 4𝑥
B. 𝑔 𝑥 = 2𝑥 3 −
′
𝑔 𝑥 = 2 3𝑥
2
𝑥2
4
+ 5𝑥
1
𝑥
2
− 2𝑥 + 5 1 = 6𝑥 − + 5
4
2
»
𝑑
𝑑𝑥
sin(𝑥) = cos 𝑥
»
𝑑
𝑑𝑥
cos(𝑥) = −sin(𝑥)
» Find the following derivatives:
A. 𝑦 = 8 cos 𝑥
𝑑𝑦
= 8 −sin(𝑥) = −8sin(𝑥)
𝑑𝑥
sin(𝑥)
2
B. 𝑓 𝑡 = 5𝑥 −
3
1
cos(𝑥)
𝑓 𝑡 = 5 2𝑥 − cos 𝑥 = 10𝑥 −
3
3
′
» We have already seen that the derivative gives the
slope at any point of the function. Slope is a measure
∆𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
of a rate of change, specifically =
.
∆𝑥
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
» Derivatives are also useful to find other rates of change,
∆𝑠
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
for instance, average velocity is =
.
∆𝑡
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
» Other applications include population growth rates,
production growth rates, water flow rates, velocity, &
acceleration.
» If a billiard ball is dropped from a height of 75
feet, its height 𝑠 at time 𝑡 is given by the
position function 𝑠 = −16𝑡 2 + 75 where 𝑠 is
measured in feet and 𝑡 is measured in seconds.
Find the average velocity over each of the
following time intervals.
A. [1, 2]
B. [1, 1.5]
C. [1, 1.1]
Find the average velocity over each of the following time intervals:
𝑠 = −16𝑡 2 + 50
A.
[1, 2]
average velocity =
B.
∆𝑠
∆𝑡
=
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
[1, 1.5]
C.
[1, 1.1]
First we must find the height of the object at these various times:
height after 1 second = 𝑠 1 = −16 1 2 + 75 = 59 ft
height after 1.1 seconds = 𝑠 1.1 = −16 1.1 2 + 75 = 55.64 ft
height after 1.5 seconds = 𝑠 1.5 = −16 1.5 2 + 75 = 39 ft
height after 2 seconds = 𝑠 2 = −16 2 2 + 75 = 11 ft
Find the average velocity over each of the following time intervals:
𝑠 = −16𝑡 2 + 50
A.
average velocity =
[1, 2]
B.
∆𝑠
∆𝑡
[1, 1.5]
=
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
C.
[1, 1.1]
Now we can calculate the average velocity over each time interval:
A.
∆𝑠
∆𝑡
B.
∆𝑠
∆𝑡
C.
∆𝑠
∆𝑡
=
𝑠 2 −𝑠(1)
2−1
=
=
𝑠 1.5 −𝑠(1)
1.5−1
=
𝑠 1.1 −𝑠(1)
1.1−1
11−59
1
= −48 ft/s
=
39−59
0.5
=
55.64−59
0.1
= −40 ft/s
= −33.6 ft/s
𝑠
𝑠
𝑠
𝑠
1 = 59 ft
1.1 = 55.64 ft
1.5 = 39 ft
2 = 11 ft
Note that this isn’t the first time you’ve seen this function. In
Algebra II and Precalculus, the function was given in a more
general form:
ℎ 𝑡 = −16𝑡 2 + 𝑣0 𝑡 + ℎ0 ,
where ℎ(𝑡) (the output) is the height of the object at any
particular time, −16 is half the acceleration due to gravity on
Earth (in ft/s/s) with the negative denoting a downward
direction, 𝑣0 is the initial velocity, and ℎ0 is the initial height.
For this problem, 𝑠 = ℎ(𝑡), 𝑣0 = 0, and ℎ0 = 75.
In Calculus I, we generalize this formula even more to:
𝑠 𝑡 =
1
𝑔𝑡 2
2
+ 𝑣0 𝑡 + 𝑠0
This time 𝑔 represents acceleration due to gravity, which is
− 32 on Earth (in ft/s/s) or −9.8 (in m/s/s), and 𝑠0 is the
initial height.
This function is called the position function and you will be
seeing it quite frequently.
» Now note this question asked for the average velocity. Pictorially
this represents the slope of the secant line passing through the
two endpoints of the interval.
tangent line
secant lines
If you wanted to find the
instantaneous velocity (or
simply velocity) of the object
at a particular time, you can
find the limit as the slope of
the secant approaches that of
the tangent, or simply find
the derivative of the function.
𝑠 𝑡 + ∆𝑡 − 𝑠 𝑡
= 𝑠′(𝑡)
∆𝑡→0
∆𝑡
𝑣 𝑡 = lim
» So, the velocity function is the derivative of the position function.
Velocity can be negative, zero, or positive. If a question asks for
the speed of an object, then we find the absolute value its velocity
because speed is always positive.
1
position = 𝑠 𝑡 = 𝑔𝑡 2 + 𝑣0 𝑡 + 𝑠0
2
𝑠 𝑡 + ∆𝑡 − 𝑠 𝑡
= 𝑠′(𝑡)
∆𝑡→0
∆𝑡
velocity = 𝑣 𝑡 = lim
speed = 𝑣(𝑡)
A. At how many points on the interval [−5, 5] is a
tangent to 𝑓(𝑥) = 𝑥 + cos(𝑥) parallel to the
secant line?
Another way this question could be worded is: “How many times is the
average rate of change on [-5, 5] equal to the instantaneous rate of change?”
Average rate of change (slope of secant):
𝑚𝑠𝑒𝑐𝑎𝑛𝑡
𝑓 5 − 𝑓(−5) 5 + cos 5 − −5 + cos(−5)
=
=
5 − −5
10
10 + cos 5 − cos(5)
=
= 1 Note: cos −5 = cos(5)
10
since it is an even function
» At how many points on the interval [−5, 5] is a
tangent to 𝑓(𝑥) = 𝑥 + cos(𝑥) parallel to the
secant line?
Average rate of change (slope of secant): 𝑚𝑠𝑒𝑐𝑎𝑛𝑡 = 1
Point on the secant line: 5, 𝑓(5) = 5, 5 + cos 5
Equation of secant line:
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
𝑦 − 5 − cos 5 = 1 𝑥 − 5
𝑦 − 5 + cos 5 = 𝑥 − 5
𝑦 = 𝑥 − cos(5)
We will now graph the function and the
secant line and compare them.
» At how many points on the interval [−5, 5] is a
tangent to 𝑓(𝑥) = 𝑥 + cos(𝑥) parallel to the
secant line?
Equation of secant line: 𝒚 = 𝒙 − 𝐜𝐨𝐬(𝟓)
The red lines represent
tangent lines parallel
to the secant. Thus,
there are 3.
B. If 𝑓 𝑥 = 2𝑥 3 − 6𝑥, at what point on the
interval 0 ≤ 𝑥 ≤ 3, if any, is the tangent to
the curve parallel to the secant line?
Similar to the last question, but must be solved without the calculator.
Also, this time it is asking for a specific point.
Average rate of change (slope of secant):
𝑚𝑠𝑒𝑐𝑎𝑛𝑡 =
𝑓
=2
3 − 𝑓(0)
=
2
3
3
−6 3
3−0
3
2
3 −6=2 3 −6=0
B. If 𝑓 𝑥 = 2𝑥 3 − 6𝑥, at what point on the
interval 0 ≤ 𝑥 ≤ 3, if any, is the tangent to
the curve parallel to the secant line?
Average rate of change (slope of secant): 𝑚𝑠𝑒𝑐𝑎𝑛𝑡 = 0
To find if a tangent to the curve is parallel to the secant without graphing, we must
find the derivative of the function and set it equal to the slope of the secant.
𝑓 ′ 𝑥 = 6𝑥 2 − 6 = 0
6𝑥 2 = 6
𝑥2 = 1
𝑥 = ±1
So, there are actually two places where
a tangent is parallel to the secant, BUT
only one of them is in the interval 0 ≤
𝑥 ≤ 3, so the answer is 1.
» At time 𝑡 = 0, a stunt diver jumps from a platform that
is 64 feet above the water. The position of the diver is
given by 𝑠 𝑡 = −16𝑡 2 + 64 where 𝑠 is measured in
feet and 𝑡 is measured in seconds.
A. When does the diver hit the water?
B. What is the diver’s velocity at impact?
» 𝑠 𝑡 = −16𝑡 2 + 64
A. When does the diver hit the water?
To answer the first question we need to set the position
function (height) equal to zero.
0 = −16𝑡 2 + 64
0 = −16 𝑡 2 − 4
0 = −16 𝑡 + 2 𝑡 − 2
𝑡 = ±2
Obviously the negative answer is extraneous, so the diver will hit the
water after 2 seconds.
» 𝑠 𝑡 = −16𝑡 2 + 64
B. What is the diver’s velocity at impact?
Here we are looking for the instantaneous velocity at the time of
impact (after 2 seconds). Thus, the derivative is necessary.
𝑠(𝑡) = −16𝑡 2 + 64
s′(𝑡) = −32𝑡
𝑣 𝑡 = −32𝑡
𝑣 2 = −32 2 = −64
So, the velocity of the diver at impact was −64 feet per second.
» Sometimes on the AP test a question will be given as a
limit, but the answer it is really looking for is the
derivative at a particular point.
» This is because the AP test really stresses the concepts
of calculus with the types of questions given, and the
derivative rules are all really just shortcuts for limits. It
is very important to recognize it when you come across
one.
A.
2+ℎ 4 −24
lim
ℎ
ℎ→0
This question is really asking:
What is the derivative of 𝑓(𝑥) = 𝑥 4 at the point where 𝑥 = 2?
Of course the trickiest part is probably recognizing the limit as a
derivative and then identifying the correct function and point.
Once that is out of the way, the question becomes easy:
𝑓 ′ 𝑥 = 4𝑥 3
𝑓 ′ 2 = 4 2 3 = 32
B.
9+ℎ−3
lim
ℎ
ℎ→0
Note this question could be rewritten as:
lim
ℎ→0
9+ℎ− 9
ℎ
This question is really asking:
What is the derivative of 𝑓(𝑥) = 𝑥 at the point where 𝑥 = 9?
𝑑 1
1
1
2
𝑥 =
𝑥 = 1=
𝑑𝑥
2 𝑥
2𝑥 2
1
1
1
𝑓′ 9 =
=
=
6
2 9 2 3
𝑓′
C.
1
1
lim
ℎ→0 ℎ 2+ℎ
−
1
2
Note this question could be rewritten as:
1
1
−2
lim 2 + ℎ
ℎ→0
ℎ
This question is really asking:
1
What is the derivative of 𝑓(𝑥) = 𝑥 at the point where 𝑥 =2?
𝑓′
𝑑 −1
1
−2
𝑥 =
𝑥 = −𝑥 = − 2
𝑑𝑥
𝑥
1
1
𝑓′ 2 = −
=−
2 2
4