Transcript Basic Differentiation Rules & Rates of Change

Basic Differentiation Rules
& Rates of Change
Chapter 3.2
Basic Differentiation Rules
• In chapter 2, you first learned how to find limits using the 𝜀−𝛿
definition
• Later, we used the definition to prove some general rules (or properties
or, more appropriately, theorems) that made finding limits much easier
• The same can be done for differentiation
• Using the definition of the derivative, we can find a few basic
theorems that make differentiation much easier
• You MUST know these rules, and the ones in the next section that
follow, by heart!!!
Theorem 3.2: The Constant Rule
THEOREM:
The derivative of a constant function, 𝑓 𝑥 = 𝑐, where c is a real number, is
𝑑
𝑐 =0
𝑑𝑥
PROOF
Let 𝑓 𝑥 = 𝑐. By the limit definition of the derivative
𝑑
𝑓 𝑥 −𝑓 𝑐
𝑐−𝑐
0
𝑐 = lim
= lim
= lim
= lim 0 = 0
Δ𝑥→0
Δ𝑥→0 Δ𝑥
Δ𝑥→0 Δ𝑥
Δ𝑥→0
𝑑𝑥
Δ𝑥
Example 1: Using the Constant Rule
𝑑𝑦
𝑑𝑥
a)
𝑦 = 7,
=0
b)
𝑓 𝑥 = 0, 𝑓 ′ 𝑥 = 0
c)
𝑠 𝑡 = −3,
d)
𝑦 = 𝑘𝜋 2 , 𝑘 is consant, 𝑦 ′ = 0
𝑑
𝑑𝑡
𝑠 𝑡
=0
Theorem 3.3: The Power Rule
THEOREM:
If n is a rational number, then the function 𝑓 𝑥 = 𝑥 𝑛 is differentiable and
𝑑 𝑛
𝑥 = 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.
1
For example, if 𝑛 = 0, then 𝑓 𝑥 = 𝑥 −1 = 𝑥 is not defined at 𝑥 = 0, so it is not differentiable at 𝑥 = 0, though
it is differentiable everywhere else.
Think of this example in terms of theorem 3.1: differentiability implies continuity if and only if discontinuity
1
implies nondifferentiability. Since 𝑥 is not defined at 𝑥 = 0, it is discontinuous at 𝑥 = 0 so it is not
differentiable at 𝑥 = 0.
Theorem 3.3: The Power Rule
THEOREM:
If n is a rational number, then the function 𝑓 𝑥 = 𝑥 𝑛 is differentiable and
𝑑 𝑛
𝑥 = 𝑛𝑥 𝑛−1
𝑑𝑥
PROOF
By the limit definition of the derivative
𝑑 𝑛
𝑥 + Δ𝑥 𝑛 − 𝑥 𝑛
𝑥 = lim
Δ𝑥→0
𝑑𝑥
Δ𝑥
Simplifying this expression requires that we know the binomial theorem, which I will use here without proof
𝑑 𝑛
𝑥 = lim
Δ𝑥→0
𝑑𝑥
𝑥𝑛
+
𝑛𝑥 𝑛−1 Δ𝑥
𝑛 𝑛 − 1 𝑥 𝑛−2 Δ𝑥 2
𝑛−1 + Δ𝑥 𝑛 − 𝑥 𝑛
+
+
⋯
+
𝑛𝑥Δ𝑥
2!
Δ𝑥
Theorem 3.3: The Power Rule
THEOREM:
If n is a rational number, then the function 𝑓 𝑥 = 𝑥 𝑛 is differentiable and
𝑑 𝑛
𝑥 = 𝑛𝑥 𝑛−1
𝑑𝑥
PROOF
𝑑 𝑛
𝑥 = lim
Δ𝑥→0
𝑑𝑥
𝑥𝑛
+
𝑛𝑥 𝑛−1 Δ𝑥
𝑛 𝑛 − 1 𝑥 𝑛−2 Δ𝑥 2
𝑛−1 + Δ𝑥 𝑛 − 𝑥 𝑛
+
+
⋯
+
𝑛𝑥Δ𝑥
2!
Δ𝑥
𝑛−2 Δ𝑥 2
𝑑 𝑛
𝑛
𝑛
−
1
𝑥
𝑥 = lim 𝑛𝑥 𝑛−1 +
+ ⋯ + 𝑛𝑥Δ𝑥 𝑛−2 + Δ𝑥 𝑛−1
Δ𝑥→0
𝑑𝑥
2!
= 𝑛𝑥 𝑛−1
Example 2: Using the Power Rule
a) 𝑓 𝑥 = 𝑥 2 , 𝑓 ′ 𝑥 = 2𝑥 2−1 = 2𝑥 1 = 2𝑥
b) 𝑔 𝑥 =
c) 𝑦 =
1
𝑥2
3
=
1
3
𝑥=𝑥 ,
−2 𝑑𝑦
𝑥 ,
𝑑𝑥
𝑔′
𝑥 =
= −2𝑥
1 1−1
𝑥3
3
−2−1
=
1 −2
𝑥3
3
= −2𝑥
−3
=
=
1
2
3𝑥 3
2
− 3
𝑥
Note that it is generally easier to rewrite radicals in their rational exponent form, and to rewrite variables in the
denominator with a negative exponent. With enough practice, you will be able to do these in your head.
Example 3: Finding the Slope of a Graph
Find the slope of the graph of 𝑓 𝑥 = 𝑥 4 when
a) 𝑥 = −1
b) 𝑥 = 0
c) 𝑥 = 1
Using the power rule we have 𝑓 ′ 𝑥 = 4𝑥 4−1 = 4𝑥 3 . The slope of the tangent line at a) 𝑥 = −1 is 𝑓 ′ −1 =
4 −1 3 = −4. The slope at b) 𝑥 = 0 is 𝑓 ′ 0 = 4 0 3 = 0. The slope at c) 𝑥 = 1 is 𝑓 ′ 1 = 4 1 3 = 4.
Example 4: Finding an Equation of a Tangent
Line
Find an equation of the tangent line to the graph of 𝑓 𝑥 = 𝑥 2 when
𝑥 = −2.
The derivative at a point is the slope of the tangent line to the curve at that point. So finding the equation of the
tangent line means using 𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1 , where (𝑥1 , 𝑦1 ) is the point on the curve. In this case, 𝑥1 = −2
so y1 = 𝑓 𝑥1 = 𝑓 −2 = −2 2 = 4. So the point is (−2,4). All that we need is the slope and this is
provided by the derivative
𝑓 ′ 𝑥 = 2𝑥 2−1 = 2𝑥
So the slope of the tangent line is 𝑓 ′ −2 = 2 −2 = −4. Therefore, the equation of the tangent line is
𝑦 − 4 = −4(𝑥 + 2)
NOTE: on the AP exam, it is never necessary to write a linear equation in 𝑦 = 𝑚𝑥 + 𝑏 form.
Theorem 3.4: The Constant Multiple Rule
THEOREM:
If f is a differentiable function and c is a real number, then cf is also differentiable and
𝑑
𝑐𝑓 𝑐 = 𝑐𝑓′(𝑥)
𝑑𝑥
PROOF
Using the derivative definition we have
𝑑
𝑐𝑓 𝑥 + Δ𝑥 − 𝑐𝑓 𝑥
𝑓 𝑥 + Δ𝑥 − 𝑓 𝑥
𝑐𝑓 𝑥 = lim
= lim 𝑐[
]
Δ𝑥→0
Δ𝑥→0
𝑑𝑥
Δ𝑥
Δ𝑥
𝑓 𝑥 + Δ𝑥 − 𝑓 𝑥
= 𝑐 lim
= 𝑐𝑓′(𝑥)
Δ𝑥→0
Δ𝑥
Example 5: Using the Constant Multiple Rule
2
𝑑𝑦
2
a) 𝑦 = 𝑥 = 2𝑥 −1 , 𝑑𝑥 = 2 −1 𝑥 −1−1 = −2𝑥 −2 = − 𝑥 2
b) 𝑓 𝑡 =
4𝑡 2
5
4
4
1
2
c) 𝑦 = 2 𝑥 = 2𝑥 ,
d) 𝑦 =
1
3
2 𝑥2
e) 𝑓 𝑥 = −
8
= 5 𝑡 2 , 𝑓 ′ 𝑡 = 5 2 𝑡 2−1 = 5 𝑡
=
3𝑥
2
𝑦′
1
2
=2
1 −2 𝑑𝑦
𝑥 3 , 𝑑𝑥
2
3
=
1
2
𝑥
1
−1
2
2
−3
𝑥
=
1
−
𝑥 2
2
−3−1
3
=
=
1
𝑥
1 −5
−3𝑥 3
=
−1
5
3𝑥 3
3
= − 2 𝑥, 𝑓 ′ 𝑥 = − 2 1 𝑥 1−1 = − 2
Example 6: Using Parentheses When
Differentiating
5
5
5
a) 𝑦 = 2𝑥 3 = 2 𝑥 −3 , 𝑦 ′ = 2 −3 𝑥 −3−1 = −
b) 𝑦 =
5
2𝑥 3
5
5
𝑑𝑦
5
=
= 8𝑥 3 = 8 𝑥 −3 , 𝑑𝑥 = 8 −3 𝑥 −3−1 = −
7
7
7
c) 𝑦 = 3𝑥 −2 = 3 𝑥 2 , 𝑦 ′ = 3 2 𝑥 2−1 =
d) 𝑦 =
15 −4
𝑥
2
7
3𝑥 −2
= 7 3𝑥
2
−15
2𝑥 4
15 −4
𝑥
8
=
−15
8𝑥 4
14𝑥
3
𝑑𝑦
= 7 9𝑥 2 = 63𝑥 2 , 𝑑𝑥 = 63 2 𝑥 2−1 = 126𝑥
Theorem 3.5: The Sum & Difference Rules
THEOREM:
The sum (difference) of two differentiable functions f and g is itself differentiable. Moreover, the derivative of
𝑓 + 𝑔 (𝑓 − 𝑔) is the sum (difference) of the derivatives of f and g.
PROOF
The proof is a direct consequence of theorem 2.2 (sum of limits).
𝑑
𝑓 𝑥 + Δ𝑥 + 𝑔 𝑥 + Δ𝑥 − 𝑓 𝑥 − 𝑔 𝑥
𝑓 𝑥 + 𝑔 𝑥 = lim
Δ𝑥→0
𝑑𝑥
Δ𝑥
= lim
Δ𝑥→0
= lim
Δ𝑥→0
𝑓 𝑥 + Δ𝑥 − 𝑓 𝑥
𝑔 𝑥 + Δ𝑥 − 𝑔 𝑥
+
Δ𝑥
Δ𝑥
𝑓 𝑥 + Δ𝑥 − 𝑓 𝑥
Δ𝑥
𝑔 𝑥 + Δ𝑥 − 𝑔 𝑥
= 𝑓 ′ 𝑥 + 𝑔′(𝑥)
Δ𝑥→0
Δ𝑥
+ lim
Example 7: Using the Sum & Difference
Rules
𝑑
a) 𝑓 𝑥 = 𝑥 3 − 4𝑥 + 5, 𝑑𝑥 𝑓 𝑥
b) 𝑔 𝑥 =
−𝑥 4
2
𝑑
𝑑
𝑑
= 𝑑𝑥 𝑥 3 + 𝑑𝑥 −4𝑥 + 𝑑𝑥 5 = 3𝑥 3 − 4 + 0 = 3𝑥 3 − 4
1
+ 3𝑥 3 − 2𝑥, 𝑔′ 𝑥 = − 2 4 𝑥 3 + 3 3 𝑥 2 − 2 = −2𝑥 3 + 9𝑥 2 − 2
Derivatives of Sine & Cosine Functions
Derivatives of Sine & Cosine Functions
Derivatives of Sine & Cosine Functions
• Recall from the previous chapter that
sin ℎ
1 − cos ℎ
lim
= 1 and lim
=0
ℎ→0 ℎ
ℎ→0
ℎ
• We will use these limits to find the derivatives of the sine and cosine
functions
• We will also use the following identities
sin 𝐴 + 𝐵 = sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
Derivatives of Sine & Cosine Functions
THEOREM:
𝑑
𝑑
sin 𝑥 = cos 𝑥 and
cos 𝑥 = − sin 𝑥
𝑑𝑥
𝑑𝑥
PROOF
From the definition
𝑑
sin 𝑥 + Δ𝑥 − sin 𝑥
sin 𝑥 cos Δ𝑥 + sin Δ𝑥 cos 𝑥 − sin 𝑥
sin 𝑥 = lim
= lim
Δ𝑥→0
Δ𝑥→0
𝑑𝑥
Δ𝑥
Δ𝑥
−sin 𝑥 1 − cos Δ𝑥
sin Δ𝑥 cos 𝑥
+ lim
Δ𝑥→0
Δ𝑥→0
Δ𝑥
Δ𝑥
= lim
1 − cos Δ𝑥
sin Δ𝑥
= − sin 𝑥 lim
+ cos 𝑥 lim
= − sin 𝑥 ⋅ 0 + cos 𝑥 ⋅ 1 = cos 𝑥
Δ𝑥→0
Δ𝑥→0 Δ𝑥
Δ𝑥
Derivatives of Sine & Cosine Functions
THEOREM:
𝑑
𝑑
sin 𝑥 = cos 𝑥 and
cos 𝑥 = − sin 𝑥
𝑑𝑥
𝑑𝑥
PROOF
From the definition
𝑑
cos 𝑥 + Δ𝑥 − cos 𝑥
cos 𝑥 cos Δ𝑥 − sin 𝑥 sin Δ𝑥 − cos 𝑥
cos 𝑥 = lim
= lim
Δ𝑥→0
Δ𝑥→0
𝑑𝑥
Δ𝑥
Δ𝑥
−cos 𝑥 1 − cos Δ𝑥
− sin Δ𝑥 sin 𝑥
+ lim
Δ𝑥→0
Δ𝑥→0
Δ𝑥
Δ𝑥
= lim
1 − cos Δ𝑥
sin Δ𝑥
= − cos 𝑥 lim
− sin 𝑥 lim
= − cos 𝑥 ⋅ 0 − sin 𝑥 ⋅ 1 = − sin 𝑥
Δ𝑥→0
Δ𝑥→0 Δ𝑥
Δ𝑥
Example 8: Derivatives Involving Sines &
Cosines
a) 𝑦 = 2 sin 𝑥, 𝑦 ′ = 2 cos 𝑥
b) 𝑦 =
sin 𝑥
2
1
2
= sin 𝑥,
𝑑𝑦
𝑑𝑥
𝑑
1
2
= cos 𝑥 =
c) 𝑓 𝑥 = 𝑥 + cos 𝑥, 𝑑𝑥 𝑓 𝑥
𝑑
cos 𝑥
2
𝑑
= 𝑑𝑥 𝑥 + 𝑑𝑥 cos 𝑥 = 1 − sin 𝑥
Derivatives of Exponential Functions
Derivatives of Exponential Functions
Theorem 3.7: Derivative of the Natural
Exponential Function
THEOREM:
𝑑 𝑥
𝑒 = 𝑒𝑥
𝑑𝑥
The natural exponent function is the unique function that is equal to its
own derivative.
Another way to think about this: At each point on the graph of the
function, the function value is the same as the slope of the tangent line
at that point.
Example 9: Derivatives of Exponential
Functions
a) 𝑓 𝑥 = 3𝑒 𝑥 , 𝑓 ′ 𝑥 = 3𝑒 𝑥
b) 𝑔 𝑥 = 𝑥 2 + 𝑒 𝑥 , 𝑔′ 𝑥 = 2𝑥 + 𝑒 𝑥
c) ℎ 𝑥 = sin 𝑥 − 𝑒 𝑥 , ℎ′ 𝑥 = cos 𝑥 − 𝑒 𝑥
Rates of Change
• The value of the derivative of a function f at a point on the graph is the
slope of the tangent line at that point
• But a slope is a rate of change of the y-values with respect to the xvalues
• We will often use this interpretation of the derivative when solving
problems
• As such, pay close attention to the units!
• A common use for rate of change is to describe the motion of a particle
in a straight line
Rates of Change
• We usually designate this motion as horizontal or vertical (with respect
to the x- and y-axes)
• If horizontal, movement to the right is considered positive while
movement to the left is negative
• If vertical, movement up is positive and movement down is negative
• A function s that gives the position of a particle (with respect to the
origin) at time t is called the position function
Rates of Change
• You should by now be familiar with the formula
𝑑
𝑑 = 𝑟𝑡 ⟺ 𝑟 =
𝑡
• If, over a period of time Δ𝑡 a particle changes its position by Δ𝑠 =
𝑠 𝑡 + Δ𝑡 − 𝑠(𝑡), then
Δs s t + Δ𝑡 − 𝑠 𝑡
=
Δ𝑡
Δ𝑡
• Note that the numerator is a change in position (hence we would use
units of length) while the denominator is a change in time (times units)
Rates of Change
𝑑
𝑡
• So is
Δ𝑠
,
Δ𝑡
each with units of change in position (or distance) over change in time
𝑑
𝑡
• Note that we use as a formula only when the rate is constant
𝑑
𝑡
• The reason is that both and
Δ𝑠
Δ𝑡
represent average velocity
𝑑
𝑡
• When the rate (or velocity) is constant, then the average is exactly
• To illustrate, suppose that you drive on a straight and level road that is 10 miles
long
• If at time 𝑡 = 0 hours your speedometer is not moving nor does it move during the
entire 20 minutes (or 1/3 hour), then you average velocity is
10 − 0
= 30 mph
1
−0
3
Rates of Change
• Now suppose that you drive from Del Rio to San Antonio, a distance
of 150 miles, and it takes you 2.5 hours
• Your average velocity is
150 − 0
= 60 mph
2.5 − 0
• But note that, on such a trip, your speedometer does move
• That is, you were not driving 60 mph during the entire 2.5 hours
• During some periods you would have been driving faster and a others
slower, or even stopped
Rates of Change
Example 10: Finding Average Velocity of a
Falling Object
If a billiard ball is dropped from a height of 100 feet, its height s at time
t is given by
𝑠 𝑡 = −16𝑡 2 + 100
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]
Example 10: Finding Average Velocity of a
Falling Object
For each interval [𝑡0 , 𝑡1 ], use the formula
a)
Δ𝑠
Δ𝑡
b)
Δ𝑠
Δ𝑡
c)
Δ𝑠
Δ𝑡
=
𝑠 2 −𝑠 1
2−1
=
𝑠 1.5 −𝑠 1
1.5−1
=
𝑠 1.1 −𝑠 1
1.1−1
=
𝑠 𝑡1 −𝑠 𝑡0
𝑡1 −𝑡0
−16 22 +100− −16 1 2 +100
=
=
2−1
=
−16 1.5 2 +100− −16 1 2 +100
1.5−1
−16 1.1 2 +100− −16 1 2 +100
1.1−1
−64+16
1
= −48 ft per sec
=
−36+16
0.5
=
−19.36+16
0.1
= −40 ft per sec
= −33.6 ft per sec
Instantaneous Rate of Change
• When velocity is constant, the graph of the position function is a
straight line (slope, which is rate of change, is a constant value)
• If the graph of the position function is not a straight line, then the
velocity differs from one instant in time to the next
• The velocity at a particular point in time is the instantaneous velocity
• The instantaneous velocity is the slope of the line tangent to the graph
at a given point
• In other words, the instantaneous velocity at a point is given by the
value of the derivative at the point
Instantaneous Rate of Change
• In general, if 𝑠 = 𝑠(𝑡) is the position function for an object moving along a
straight line, the velocity of the object at time t is
𝑠 𝑡 + Δ𝑡 − 𝑠 𝑡
𝑣 𝑡 = 𝑡 = lim
Δ𝑡→0
Δ𝑡
Velocity is a vector function so it can be positive, negative, or zero
Speed is the absolute value of velocity
The position function for a free-falling object is given by
1 2
𝑠 𝑡 = 𝑔𝑡 + 𝑣0 𝑡 + 𝑠0
2
Here, g is acceleration due to gravity, 𝑣0 is the initial velocity and 𝑠0 is the
initial position
𝑠′
•
•
•
•
Example 11: Using the Derivative to Find
Velocity
At time 𝑡 = 0, a diver jumps from a platform diving board that is 32 feet
above the water. The position of the diver is given by
𝑠 𝑡 = −16𝑡 2 + 16𝑡 + 32
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?
Example 11: Using the Derivative to Find
Velocity
For part a), the position function, with 𝑠 = 0, will allow you to solve for t
0 = −16𝑡 2 + 16𝑡 + 32
0 = 𝑡 2 − 𝑡 − 2 = (𝑡 − 2)(𝑡 + 1)
So 𝑡 = 2 or 𝑡 = −1. Only the positive value has meaning here, so the diver reaches the water after 2 seconds.
The velocity at impact is the derivative of s at time 𝑡 = 2 seconds
𝑣 𝑡 = 𝑠 ′ 𝑡 = −32𝑡 + 16
𝑣 2 = 𝑠 ′ 2 = −32 2 + 16 = −48 ft per sec
Exercise 3.2a
• Page 136, #1-30
Exercise 3.2b
• Page 136, #31-61 odds, 63-66, 73-76
Exercise 3.2c
• Page 138, #89-100, 103-115 odds