chap3 notes
TRANSCRIPT
Lirong
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Chapter 3 The derivative
3.1 The Derivative and Rates of Change
Definition: The derivative of a function f is the function f ′ defined by
f ′(x) = limh→0
f(x + h)− f(x)
h
for all x where the limit exists.
(insert graph of the above idea)
Remark: Geometric Interpretation of the derivativeNote that let x = a, in the above definition, we then have
f ′(a) = limh→0
f(a + h)− f(a)
h
i.e. f ′(a) is the slope of the tangent line to the graph of f at the point (a, f(a)), Soy − f(a) = f ′(a) ∗ (x− a) is the equation of the tangent line. Here ”a” is x coordinateof the tangent point.
Differentiating a given function f by direct evaluation of the limit in the definitioninvolves carrying out four steps:
1). Write the definition equation of the derivative.2).Substitute the expression f(x+h) and f(x) as determined by the perticular functionf.3). Simplify the result by algebraic methods until it is possible to ..4). Apply appropriate limit laws to finally evaluate the limit.
Example 1 Let f(x) =x
x + 3,
(a) Compute f ′(x)(b) Find the equation of the tangent line at (0,f(0)).
Remark Even when the function f is rather simple, this four-step process for computingf ′ directly from the definition of the derivative can be time consuming. Also, Step 3 mayrequire considerable ingenuity. Moreover, it would be very repetitious to continue torely on this process. To avoid tedium, we want a fast, easy, short method for computingf ′.
This is the focus of this chapter: the development of systematic methods (”rules”)for differentiating those functions that occur most frequently. Such functions includepolynomials, rational functions, the trigonometric functions sin x and cos x, and combi-nations of such functions.
Rule If f(x) = ax2 + bx + c, Then
f ′(x) = 2ax + b
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Proof. (given in section 2.1)
Example 2a) f(x) = 3x2 − 4x + 5, f ′(x)=?b) g(t) = 2t− 5t2, g′(t)=?
(relation between the graph of f and f ′) see homework P117 30-35.(see also Example 8 in textbook)
Differential Notation
An important alternative notation is to write ∆x in place of h, and
∆y = f(x + ∆x)− f(x)
for the resulting change in y. The The differential quotientf(x + h)− f(x)
h=
∆y
∆x,
where the ∆y means the change in y, and the ∆x means the change in x. So
f ′(x) = limh→0
f(x + h)− f(x)
h= lim
∆x→0
∆y
∆x=
dy
dx
and f ′(a) =dy
dx|x=a
Example 2 (cont.) try these notation
a)dy
dx= 6x− 4, so
dy
dx|x=1 = 2
b) denote z = 2t− 5t2,dz
dt= 2− 10t,
dz
dt|t=1 = −8
Remark The benefit of using the Differential Notation is that they mark out which theindependent variable, which is the dependent variable.f(x), f ′(x): f is the function name and f ′ denote its derivative.dy
dx: Differential Notation. As x changes, y changes. x is independent variable, y is
dependent on x.
f ′(x) anddy
dxare used almost interchangeably in mathematics, so we need to be familiar
with both version of notation.
We also need to know thatdy
dxis a simbol representing the derivative. It is not the
quotient of two seperate quantities.
(relation between the graph of f and f ′) see homework P117 30-35.(see also Example 8 in textbook)
Core Problems
P116: 5, 3, 9, 11, 15, 17, 19,30-35.
Find the equation of the tangent lines at (1,f(1)) for 11, 15,17,19.
Rates of change
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Suppose that Q is a quantity that varies with time t, and write Q = f(t) for the valueof Q at time t. For example, Q might beThe size of a population (such as kangaroos, people, or bacterria)The distance traveled t hours after the beginning of the journeyThe volume of a balloon being inflated(more in the book)
Definition(1) The average rate of change of Q from time t to t + ∆t is the ratio
∆Q
∆t=
f(t + ∆t)− f(t)
∆t
(2) The (instantaneous) rate of change of Q at time t is
lim∆t→0
∆Q
∆t= lim
∆t→0
f(t + ∆t)− f(t)
∆t
Observe that the left hand side of the equation is exactly f ′(t). Hence the instantaneousrate of change of Q = f(t) is
dQ
dt= f ′(t)
Remark: additional important conclusionBecause a positive slope corresponds to a rising tangent line and a negative slope cor-responds to a falling tangent line, we say that
Q is increasing at time t ifdQ
dt> 0
Q is decreasing at time t ifdQ
dt< 0
Example 1 A cylindrical tank has a vertical axis and is initially filled with 600 gal ofwater. The tank takes 60 min to empty after a drain in its bottom is opened. Supposethat the drain is opened at time t = 0. Suppose that the volume V of the water remainingin the tank after t mimutes is
V (t) =1
6(60− t)2 = 600− 20t +
1
6t2
gallons.Part1: Find the instantaneous rate at which water is flowing out of the tank at timet = 15(min) and at time t = 45(min)Part2: Find the average rate at which water flows out of the tank during the half hourfrom t = 15 and t = 45.
Sol: The instantaneous rate of change of the volume V (t) of water in the tank is given
by the derivativedV
dt= −20 +
1
3t
At the instant t = 15 and t = 45 we obtain
V ′(15) = −20 +1
3· 15 = −15
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V ′(45) = −20 +1
3· 45 = −5
The average rate of change of the volume V (t) is
∆V
∆t=
V (45)− V (15)
45− 15=
16(60− 45)2 − 1
6(60− 15)2
45− 15=−300
30
Remark: note that all these rate of change are negative. This is consistent with thefact that V is a decreasing function of t
Velocity(1) position functionSuppose that a moving particle’s location x at time t is given by a position functionx = f(t).(2) average velocity from time a to b
∆x
∆t=
f(b)− f(a)
b− a
average velocity from time t to t + ∆t
∆x
∆t=
f(t + ∆t)− f(t)
∆t
(3)(instantaneous) velocity at time t
v(t) = lim∆t→0
∆x
∆t= lim
∆t→0
f(t + ∆t)− f(t)
∆t
Notice that the limit on the right in the above equation is the definition of the derivativeof t at time t. Therefore,
v(t) =dx
dt= f ′(t).
Velocity is the instantaneous rate of change of position.
Example 2 If a ball is thrown vertically upward from the roof of 64 foot building witha velocity of 96 ft/sec, its height after t seconds is s(t) = 64 + 96t − 16t2. What is themaximum height the ball reaches? What is the velocity of the ball when it hits theground (height 0)?
Remark: at the beginning, when t = 0, the height function s(t), which is the positionfunction is increasing, and its velocity function v(t) = s′(t) = 96 − 32t is positive Astime t grow bigger, s(t) will reach its highest point. Then, it will become a decreasingfunction, its derivative v(t) will become negative. Since v(t) is a continuous function,it will not jump from positive to negative. At certain time t, we will have v(t) = 0 andthat is when s(t) reach its biggest value.
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Sol:The instantaneous velocity at time t of the ball is v(t) = s′(t) = 96− 32t. when v(t) = 0the ball will reach its maximum height. So solve 96− 32t = 0 will give the time t.
96− 32t = 0−32t = −96
t = 3
since s(t) = 64 + 96t− 16t2 and the maximum value of s(t) is when t = 3 we know themaximum value is s(3) = 64 + 96 · 3− 16 · 32 = 208.
When the ball hits the ground, the height function (position function) s(t) = 0, i.e.64 + 96t− 16t2 = 0, solve this for t.
64 + 96t− 16t2 = 0−16t2 + 96t + 64 = 0
t2 − 6t− 4 = 0
t = 6±√
36+4·1·42
t = 6+√
36+4·1·42
since t > 0
t = 6+√
522
The velocity at this moment is v(6+√
522
) = 96− 326+√
522
= −16√
52
Remark: Negative velocity means the direction of the ball is downward when it hits theground.
Core problems P118: (note the hint before problem 44) 41,42(sketch graph),52,53P117: 21,23,25,27,29,39.some hints:42: P (t) = 100(1 + 0.3t + 0.04t2).”How long does it take for this population to double its initial size” is asking when t=?will P (t) = 2P (0). That is we need to solve 100(1 + 0.3t + 0.04t2) = 2 · 100.39: position function x(t) = 100t− 5t2. so the velocity function is dx
dt= 100− 10t. When
the car comes to a stop, the velocity should be 0.
General Rates of Changes
If we have Q = f(t), then the average rate of change from time t to t + ∆t is the ratio
∆Q
∆t=
f(t + ∆t)− f(t)
∆t
and the (instantaneous) rate of change of Q at time t is
lim∆t→0
∆Q
∆t= lim
∆t→0
f(t + ∆t)− f(t)
∆t=
dQ
dt
If we have y = f(x), the average rate of change with respect to x from x to x + ∆x isthe ratio
∆y
∆x=
f(x + ∆x)− f(t)
∆x
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and the (instantaneous) rate of change of y at x is
lim∆x→0
∆y
∆x= lim
∆x→0
f(x + ∆x)− f(x)
∆x=
dy
dx
Remark: Notice that we added the ”with respect to”. Very often the independentvariable in the problem is not time t, or x.
Example 7(a) Find the rate of change of the area A of a square with respect to itsedge length x.
Sol: Since A = x2, the rate of change A with respect to x is
dA
dx= 2x.
Example 7(b) (Continue the problem in Example 7(a).) If the edge length x is changingwith time t, say x = 2t + 3. Find the rate of change of A with respect to t at t = 1.
Sol: If we want to find the rate of change of A with respect to t, we need to find therelationship between A and t.Since A = x2, and x = 2t + 3, we then have A = (2t + 3)3 = 4t2 + 12t + 9.So,
dA
dt= 8t + 12
anddA
dt|t=1= 8 + 12 = 20.
P118: 36 The Celsius temperature C is given in terms of the Fahrenheit temperatureF by C = 5
9(F − 32). Find the rate of change of C with respect to F and the rate of
change of F with respect to C.
Core problems P118: 37,44(note hint), 45,47, 50,51.
3.2 Basic Differentiation Rules
Definition
f is differentiable at x if f ′(x) exists.f is differentiable if f is differentiable at each point of its domain.
Notation If y = f(x) then
y′ = f ′(x) =dy
dx=
df(x)
dx= Dxf(x)
The Differentiation Rules
1)The constant rule: The derivative of a constant
If f(x) = c (a constant) for all x, then f ′(x) = 0.
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For example, if f(x) = 9, then f ′(x) = 0
2)The power rule: The derivative of a power function
If f(x) = xn, then f ′(x) = nxn−1.
For example, if f(x) = x, then f ′(x) = 1If f(x) = x2, f ′(x) = 2x.If f(x) = x3, f ′(x) = 3x2
If f(x) =√
x(= x12 ), f ′(x) = 1
2x−
12 which is equivalent to f(x) = 1
2√
x
3) The derivative of a linear combination
A linear combination of the functions f and g is a function of the form af + bg where aand b are constants. For example, 3f + 8g is a linear combination of f and g. 9g + 10fis also a linear combination of f and g.
If we know the limit of f(x) and g(x) at point x = c, then we can find the limit ofaf(x) + bg(x) at the point x = c. since
limx→c
af(x) + bg(x) = a limx→c
f(x) + b limx→c
g(x)
This is called the linear property of the limit operation. Recall that the derivative isalso defined by limits.
If h(x) = af(x) + bg(x), where a and b are fixed real numbers, then
h′(x) = af ′(x) + bg′(x)
For example, we know if we have f(x) = x4, g(x) = x2,then (by the power rule)f ′(x) = 4x3, g′(x) = 2x.So if we let h(x) = x4 + x2, h′(x) = ? or p(x) = 3x4 + 7x2, p′(x) =?Notice h(x) is a linear combination of f(x) and g(x), we know its derivative is also alinear combination of f ′(x) and g′(x).h′(x) = f ′(x) + g′(x) = 4x3 + 2x. andp′(x) = 3f ′(x) + 7g′(x) = 3 · 4x3 + 7 · 2x = 12x3 + 14x
Question: what is b = 0 in the linear combination af + bg, what is the derivativeof af(x) ?4) The derivative of a polynomial functionThe above examples are actually both polynomials. To sum up our finding, we see
Letp(x) = anx
n + an−1xn−1 + · · ·+ a2x
2 + a1x + a0
thenp′(x) = nanx
n−1 + (n− 1)an−1xn−2 + · · ·+ 2a2x + a1
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For example, let p(x) = 2x3 − 7x2 + 3x + 4,its derivative is p′(x) = 6x2 − 14x + 3.
Example4. Write an equation for the straight line that is tangent to the graph ofy = 2x3 − 7x2 + 3x + 4 at the point (1,2).
Sol. First we compute the derivative
dy
dx= 6x2 − 14x + 3.
At point (1,2), the tangent line has slope dydx|x=1 = 6− 14 + 3 = −5.
With the point (1,2) and slope -5, we can find point-slope formula of the equation ofthe tangent line: y − 2 = −5(x− 1), i.e. y = −5x + 7.
5)Product rule: The derivative of the product of two functions
Let h(x) = f(x)g(x), then h′(x) = f ′(x)g(x) + f(x)g′(x).
To write it more briefly, (fg)′ = f ′g + fg′.
Example 6. Find the derivative of f(x) = (1 − 4x3)(3x2 − 5x + 2) without firstmultiplying out the two factors.
Sol.f ′(x) = (1− 4x3)′(3x2 − 5x + 2) + (1− 4x3)(3x2 − 5x + 2)′
= (−12x2)(3x2 − 5x + 2) + (1− 4x3)(6x− 5)= −60x4 + 80x3 − 24x2 + 6x− 5
6)the Quotient rule: the derivative of the quotient of two functions
If f and g are differentiable at x and f(x) 6= 0, then g/f is differentiable at x and
(g(x)
f(x))′ =
g′(x)f(x)− g(x)f ′(x)
[f(x)]2
Reciprocal rule: the derivative of the reciprocal of a function.
If f is differentiable at x and f(x) 6= 0, then (1
f(x))′ = − f ′(x)
[f(x)]2
Example 9. Find z′(t) =dz
dtwhere z is given by z =
1− t3
1 + t4
Sol.dz
dt=
(1− t3)′(1 + t4)− (1− t3)(1 + t4)′
(1 + t4)2
=(−3t2)(1 + t4)− (1− t3)(4t3)
(1 + t4)2
=t6 − 4t3 − 3t2
(1 + t4)2
Theorem 5 Power rule for a negative integer n.
If n is a negative integer, then Dxxn = nxn−1.
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Example 8
Dx
(5x4−6x+7
2x2
)= Dx(
52x2 − 3x−1 + 7
2x−2)
= 52(2x)− 3(−x−2) + 7
2(−2x−3) = 5x + 3
x2 − 7x3
To sum up,1)The constant rule: If f(x) = c (a constant) for all x, then f ′(x) = 0.2)The power rule: If f(x) = xn, then f ′(x) = nxn−1.3)The derivative of a linear combination If h(x) = af(x) + bg(x), where a and b arefixed real numbers, then h′(x) = af ′(x) + bg′(x)
4) The derivative of a polynomial function Let p(x) = anxn+an−1x
n−1+...+a2x
2+a1x+a0
then p′(x) = nanxn−1 + (n− 1)an−1x
n−2 +... + 2a2x + a1
5)Product rule: (fg)′ = f ′g + fg′.6)the Quotient rule: If f and g are differentiable at x and f(x) 6= 0, then g/f is differ-
entiable at x and (g(x)
f(x))′ =
g′(x)f(x)− g(x)f ′(x)
[f(x)]2
Core Problem: P128: 1,3,13,19,35,27,31,15, 43,45.
Proof for the basic differentiation rulesNow let’s prove these basic differentiation rules:1)The constant rule: If f(x) = c (a constant) for all x, then f ′(x) = 0.
Proof. Because f(x + h) = c, f(x) = c, we have
f ′(x) = limh→0
f(x + h)− f(x)
h= lim
h→0
c− c
h= lim
h→0
0
h= 0
2)The power rule: If f(x) = xn, then f ′(x) = nxn−1.
Proof. For a positive integer n, the identity
bn − an = (b− a)(bn−1 + bn−2a + bn−3a2 + · · ·+ ban−2 + an−1)
is easy to verify by multiplication. Thus if b 6= a, then
bn − an
(b− a)= bn−1 + bn−2a + bn−3a2 + · · ·+ ban−2 + an−1
Because each of the n terms on the right-hand side approaches an−1 as b → a, this tellsus that
limb→a
bn − an
b− a= nan−1
(apply power law, sum law of limits). Now let b = x + h and a = x, so that h = b− a.Then h → 0 as b → a. and hence
f ′(x) = limh→0
(x + h)n − xn
h= nxn−1
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This proves the power rule.
3)The derivative of a linear combination will be left as a homework problem.
4) The derivative of a polynomial functionis a direct conclusion from 1) 2) 3).
5)Product rule: suppose f and g are both differentiable functions, and let h(x) =f(x)g(x) then h′(x) = f ′(x)g(x) + f(x)g′(x).
Proof. We use an ”add and subtract” device.
h′(x) = limh→0
h(x + h)− h(x)
h
= limh→0
f(x + h)g(x + h)− f(x)g(x)
h
= limh→0
f(x + h)g(x + h)− f(x)g(x + h) + f(x)g(x + h)− f(x)g(x)
h
= limh→0
f(x + h)g(x + h)− f(x)g(x + h)
h+ lim
h→0
f(x)g(x + h)− f(x)g(x)
h
= limh→0
[f(x + h)− f(x)]g(x + h)
h+ lim
h→0
f(x)[g(x + h)− g(x)]
h
= limh→0
f(x + h)− f(x)
hlimh→0
g(x + h) + f(x) limh→0
g(x + h)− g(x)
h= f ′(x)g(x) + f(x)g′(x)
In this proof we used the sum law and product law for limits, the definition of f ′(x) andg′(x), and the fact
limh→0
g(x + h) = g(x).
This equation is true because g is differentiable.
Remark Differentiable =⇒ continuous. The converse is not true.
6)Reciprocal rule:
If f is differentiable at x and f(x) 6= 0, then (1
f(x))′ = − f ′(x)
[f(x)]2
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Proof. Let h(x) =1
f(x), then
h′(x) = limh→0
h(x + h)− h(x)
h
= limh→0
1
h
(1
f(x + h)− 1
f(x)
)= lim
h→0
1
h
f(x)− f(x + h)
f(x + h)f(x)
= limh→0
f(x)− f(x + h)
h
1
f(x + h)f(x)
= limh→0
f(x)− f(x + h)
hlimh→0
1
f(x + h)f(x)
= − limh→0
f(x + h)− f(x)
hlimh→0
1
f(x + h)f(x)
= −f ′(x)1
[f(x)]2
The Quotient rule: If f and g are differentiable at x and f(x) 6= 0, then g/f is differen-
tiable at x and (g(x)
f(x))′ =
g′(x)f(x)− g(x)f ′(x)
[f(x)]2
Proof. We apply the product rule to the factorization
g(x)
f(x)= g(x) · 1
f(x)
This gives
(g(x)
f(x))′ = g(x)(
1
f(x))′ + g′(x)
1
f(x)= g(x)[−f ′(x)
1
[f(x)]2] + g′(x)
1
f(x)
=−g(x)f ′(x)
[f(x)]2+
g′(x)
f(x)
=−g(x)f ′(x)
[f(x)]2+
g′(x)f(x)
[f(x)]2
=g′(x)f(x)− g(x)f ′(x)
[f(x)]2
The extended product ruleSuppose function f, g and h are differentiable, then(f(x)g(x)h(x))′ = f ′(x)g(x)h(x) + f(x)g′(x)h(x) + f(x)g(x)h′(x)
Suppose function f, g, p and q are differentiable, then
(f(x)g(x)p(x)q(x))′ = f ′(x)g(x)p(x)q(x) + f(x)g′(x)p(x)q(x)+ f(x)g(x)p′(x)q(x) + f(x)g(x)p(x)q′(x)
Similar results can be derived in the same pattern when more function are being multi-plied.
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P123: 17 can use extended product rule. g(y) = 2y(3y2 − 1)(y2 + 2y + 3).
More exampleIf f is the focal length of a convex lens and an object is placed at a distance p from thelens, then its image will be at a distance q from the lens, where f , p, and q are relatedby the lens equation
1
f=
1
p+
1
q
Suppose that f = 3. What is the rate of change of p with respect to q when q = 6?
Sol. With f = 3 the above relation becomes
1
3=
1
p+
1
q
Obviously, when q changes, p has to change accordingly. Since we need to find the rate
of change of p with respect to q, i.e.dp
dq, we have to express p in terms of q, i.e. find the
function f such that p = f(q), thendp
dq= f ′(q).
Since1
3=
1
p+
1
qwe have
1
p=
1
3− 1
q=
q − 3
3qHence p =
3q
q − 3.
Then by quotient rule
dp
dq=
(q − 3)(3q)′ − (q − 3)′(3q)
(q − 3)2=
(q − 3)3− (3q)
(q − 3)2=
−9
(q − 3)2
Sodp
dq|q=6 =
−9
(6− 3)2= −1
Core problems
P129: 55,56,59,51, 53.
P128: Use the extended product rule to do problem 5,7.
Prove the linear combination law of the derivative.
3.3 The Chain Rule
Section 3.3 The Chain Rule
Review of Composition of functions
Before we start to discuss the Chain Rule for computing derivatives, we need to reviewthe concept of composition of functions. We have discussed this topic in the limitcomputation. Let’s see a few examples.
h(x) =√
x2 + 1h(x) = ln(x2 + 1)h(x) = sin(x2 + 1)
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Besides sum, substraction,multiplication, division, composition of function is anotherway of combining known functions to create new functions. The above function are allcreated by one ”outer” function and one ”inner” function.
Take the h(x) =√
x2 + 1 as an example, the ”outer” function is f(x) =√
x, the innerfunction g(x) = x2 + 1.
h(x) = f(g(x))
Chain Rule
We will need the Chain Rule to find the derivative of a composition of functions. Simplesubstitution will not work.f(x) =
√x h(x) =
√x2 + 1
f ′(x) =1
2√
xh′(x) 6= 1
2√
x2 + 1
Theorem
If f and g are differentiable functions, then
(f(g(x)))′ = f ′(g(x)) · g′(x)
Let see how to use this theorem to find the derivative of a compostion function.Example 1Let h(x) = (x3 − x2 + 1)5, find f ′(x). (P132:1-12)
Sol.First, identify the ”outer” function and ”inner” function.h(x) = f(g(x)) where f(x) = x5 (”outer” function) and g(x) = x3 − x2 + 1.(”inner”function)
Second, find the related terms in the theorem.since f(x) = x5
f ′(x) = 5x4
f ′(g(x)) = 5(g(x))4 = 5(x3 − x2 + 1)4
since g(x) = x3 − x2 + 1.g′(x) = 3x2 − 2x.
Third, apply the theorem: h′(x) = (f(g(x)))′ = f ′(g(x))·g′(x) = 5(x3−x2+1)4·(3x2−2x)(remember to put brackets at the proper places)
Corollary 1Let u = g(x), then
d
dxf(u) =
d
duf(u)
du
dx
Theorem (Generalized Power Rule)
d
dx(un) = nun−1du
dx
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where n is an integer.
Proof. This is an application of Corollary 1. Let f(u) = un, then
d
dx(un) =
d
du(un)
du
dx= nun−1du
dx
Example 1(redo)sol. Let h(x) = u5, u = x3 − x2 + 1. then apply Generalized Power Rule:
h′(x) =d
dxh(x) =
d
dx(u5) = 5u4du
dx
now u = x3 − x2 + 1, dudx
= 3x2 − 2x, hence
h′(x) = 5(x3 − x2 + 1)4(3x2 − 2x)
more example
Exd
dx
1
(x3 − x2 + 1)3
Exd
dx
x + 1
(x2 + 5)3
Corollary 2Let y = f(u) and u = g(x), then
dy
dx=
dy
du
du
dx
Example 2
Let y = u3 + 1 and u = x2 + x + 1, use chain rule to computedy
dxand express your
answer in terms of x. (P132: 13-20)
sol. By Corollary 2,
dy
dx=
dy
du
du
dx=
d
du(u3 + 3)
d
dx(x2 + x + 1)
= 3u2(2x + 1)= 3(x2 + x + 1)2(2x + 1)
Remark1. express your final answer in terms of x. Do not leave any u in your final answer.2. notice the proper use of round brackets. It is still correct if you did not expand theexpression such as 3(x2 + x + 1)2 · (2x + 1), but 3(x2 + x + 1)2 · 2x + 1 is definitely anwrong answer.
more example Find derivative for the following functions:
1) y =1
(2x3 − x + 7)2
2) h(z) = (z − 1
z + 1)5
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Core problemsP132: 1,4,7,9,11,13,17,21,25,29
Rate of Change(application of chain rule) Suppose that a quantity p depends on thequantity q, which in turn depends on quantity t.
dependent variable p, intermediate variable q, independent variable t.
Then the derivatives that appear in the chain rule
dp
dt=
dp
dq
dq
dt
are rates of changes of these variables with respect to one another.
Example 6A spherical balloon is being inflated. The radius r of the balloon is increasing at therate of 0.2 cm/s when r=5cm. At what rate is the volume V of the balloon increasingat that instant?
sol. Givendr
dt= 0.2 when r = 5cm, we want to find
dV
dtat that instant.
V =4
3πr3
We see that dVdr
= 4πr2. so the chain rule gives
dV
dt=
dV
dt
dr
dt= 4πr2dr
dt
Now, when r = 5, 4πr2 = 4π52 anddr
dt= 0.2 therefore
dV
dt= 4π52(0.2) ≈ 62.83 (cm3/s)
RemarkThe key in finding the correct answer for this example is to translate the word intoproper derivative equation.
ExampleImagine a spherical raindrop is falling through water vapor in the air. Suppose that thevapor adheres to the surface of the raindrop in such a way that the time rate of increaseof the mass M of the droplet is proportional to the surface area S of the droplet. If theinitial radius of the droplet is, in effect, zero and the radius is r = 1mm after 20s, whenis the radius 3mm?(given: M = 4
3πρr3 and S = 4πr2 where ρ is the density of water.)
Sol. We are given the information that dMdt
is proportional to S. Let suppose the ratiois a constant number k.
dM
dt= kS = k(4πr2)
Since M = 43πρr3 we have
dM
dr= 4πρr2
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By chain ruledM
dt=
dM
dr
dr
dt= 4πρr2dr
dt
compare the above equation for dMdt
, we must have
k(4πr2) = 4πρr2dr
dt
hencedr
dt=
k
ρ
The rate of change of r with respect to t is constant. Now it takes 20s for r to changefrom 0 to 1mm, means that
∆r
∆t=
1− 0
20= 0.05
suppose after t seconds, the radius is 3mm. we must have:
3− 0
t= 0.05
Solving this gives us t = 60. It takes 1 minute for r to grow to 3mm.
Core problemsP132: 49,51,53,55,57,59
webwork:13: x is the ”supply”, p is the ”price”14 taking derivative of A w.r.t. r. then calculate according to different value of r
3.4 Derivatives of Algebraic Functions
Section 3.4 Derivatives of Algebraic Functions
Remark: this section, we will be learning to differentiate functions with rational expo-nents.
Recall The power rule: If f(x) = xn, then f ′(x) = nxn−1.
Proof. For a positive integer n, the identity
bn − an = (b− a)(bn−1 + bn−2a + bn−3a2 + · · ·+ ban−2 + an−1)
is easy to verify by multiplication. Thus if b 6= a, then
bn − an
(b− a)= bn−1 + bn−2a + bn−3a2 + · · ·+ ban−2 + an−1
Because each of the n terms on the right-hand side approaches an−1 as b → a, this tellsus that
limb→a
bn − an
b− a= nan−1
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(apply power law, sum law of limits). Now let b = x + h and a = x, so that h = b− a.Then h → 0 as b → a. and hence
f ′(x) = limh→0
(x + h)n − xn
h= nxn−1
This proves the power rule.
Now what if the exponent is a negative integer? Say, n = −m, where m is an positiveinteger. Now f(x) = xn, hence,
f(x) = x−m =1
m.
By quotient rule,
f ′(x) = ( 1xm )′ = −(xm)′
(xm)2= −mxm−1
x2m = −mx(m−1)−2m
= −mx−m−1 = nxn−1
This proves the power rule is true even when the exponent is negative integer, hence itis true for all integer n.
Now let extend the result even more by proving that it is still true when the exponentis a rational number, say, r = p/q. Here, p and q are integers, and q 6= 0.
Recall that rational powers are defined in terms of integeral roots and powers as follows:
xp/q = q√
xp = ( q√
x)p
In problems 72 through 75 (page 141), using the similar identity as above, it is provedthat
( q√
x)′ = (x1/q)′ =1
qx
1q−1 =
1
qx−
q−1q
Now, to find the derivative of f(x) = ( q√
x)p, we need the Chain Rule. Let set up anintermediate variable u = q
√x, then y = f(x) = up.
dy
dx=
dy
du
du
dx=
d
du(up)
d
dx( q√
x) = pup−1 1
qx−
q−1q
= p( q√
x)p−1 1qx−
q−1q
= pxp−1
q 1qx−
q−1q
= pqx
p−1q− q−1
q
= pqxp/q−1
This proves (xp/q) = pqxp/q−1.
Now we have shown that the power rule (xr)′ = rxr−1 even when r is a rational number.
Example 1(a) d
dx
√x = (x1/2)′ = 1
2x−1/2 = 1
2√
x
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(b) If y =√
x3, then dydx
= (x3/2) = 32x1/2 = 3
2
√x
(c) If g(t) = 13√
t2= t−2/3, then g′(t) = −2
3t−5/3 = − 2
33√
t5
Remark:In parts (a) and (b), due to the squre root, we must have x ≥ 0.
Moreover, in part (a), we see that x 6= 0, otherwise the derivative function will involvedivision by 0. (x = 0 is in the domain of
√x, but not in the domain of the derivative of√
x)
Generalized power ruleRecall the generalized power rule from section 3.3
d
dx(un) = nun−1du
dx
where n is an integer. Now, for a more general form, let y = ur and r = p/q is rational.Then
dy
du= rur−1
thendy
dx=
dy
du
du
dx= rur−1du
dx
Suppose u = f(x), this is the Generalized power rule
d
dx[f(x)]r = r[f(x)]r−1 · f ′(x)
Example 2
ddx
[√
4− x2] = ddx
[(4− x2)1/2] = 12(4− x2)−1/2 · (4− x2)′
= 12(4− x2)−1/2 · (−2x)
= −x(4− x2)−1/2
= −x√4−x2
except where x = 2 or − 2 (division by 0) or where |x| > 2 (square root of a negativenumber). Therefore, the above equation hold only if −2 < x < 2.
RemarkNote that if f(x) =√
4− x2, we have f ′(x) =−x√4− x2
,
(1) The domain of f(x) is [−2, 2], while the domain of f ′(x) is (-2,2). Which means: atsome of the x values, f is continuous AND differentiable; at some of the x values, f iscontinuous but NOT differentiable. Differentiability implies continuity. The converse isnot true.
(2) Note f ′(x) →∞ as x → −2+ and f ′(x) → −∞ as x → +2−.
Example 3If y = 5
√x3 − 2
3√
x, then
y = 5x3/2 − 2x−1/3
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sody
dx= 5 · (3
2x1/2)− 2 · (−1
3x−4/3) =
15
2x1/2 +
2
3x−4/3 =
15
2
√x +
2
33√
x4
Core problemsP144: 1,9,19,33,39,45, 4963,65.
webwork:
6. The volume of a sphere with radius r is V = 43πr3. It’s surface area is A = 4πr2.
Suppose, now that such a sphere is growing or shrinking. Find the rate of change of thevolume with respect to its surface area when r = 75.
What is dVdA
?
Hint: One way to compute dVdA
is to first solve for V in terms of A.
7. There are two points on the circle x2 + y2 = 1 at which the tangent line has slope -4.What are they?
The point on the upper half of the circle is: ( , )
The point on the lower half of the circle is: ( , )
In fact, for any slope m, there are two points on that circle at which the slope of thetangent line is m.
What is the point on the upper half? ( , )
Section 3.5 Maxima and Minima of Functions on Closed Intervals
Remark: Introducing the theory of finding maxes and mins on a closed interval forcontinuous functions.
DefinitionIf c is in the interval [a,b], then f(c) is called the minimum value of f(x) on [a,b] if
f(c) ≤ f(x) for all x in [a, b]
If d is in the interval [a,b], then f(d) is called the maximum value of f(x) on [a,b] if
f(d) ≥ f(x) for all x in [a, b]
Theorem 1If f(x) is continuous on [a,b], then f has maximum value and minimum value on [a,b].
i.e. there exist number c and d in [a,b] such that f(c) is the minimum value and f(d) isthe maximum value.
Theorem 2If (1) f(x) is continuous on [a,b],
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(2) f(c) is a maximum or minimum value of f, and(3) f ′(c) exists, then
f ′(c) = 0
Beware The converse of Theorem B is false. That is, the fact that f ′(c) = 0 is notenough to imply that f(c) is a maxima or minima. For example, for function f(x) = x3
we have f ′(x) = 3x2. Then f ′(0) = 0. However, f(0) is not a maxima or a minima.(draw the graph)
Brief proof given in class. Detailed proof is on textbook page 143.
Definitionx = c is a critical point of f if c is in the domain of f and(1) f ′(c) = 0 or(2) f ′(c) DNE
Theorem 3If f is continuous on [a, b], and the only critical points that f has are c1, c2, · · · , ck suchthat a < c1 < c2 < · · · < ck < b. Then in the following list:
f(a), f(c1), f(c2), · · · , f(ck), f(b)
the largest value is the maximum value of f on [a,b] and the smallest value is the mini-mum value of f on [a,b]
STEPS to find max and min1) Know the function f and check if f is continuous on a closed interval.2) If 1) is true, find the critical points of f ( f ′(c) = 0 or f ′(c) DNE)3) evaluate f at the critical points and the endpoints of the closed interval [a,b]. Com-pare to find the max and min.
Example 4Find the max and min of function f(x) = 2x3 − 6x on interval [0,2]
1) f is a polynomial which is continuous everywhere, hence it is continuous on [0,2].2)To find the critical points of f we need to calculate f ′:
f ′(x) = 6x2 − 6 = 6(x2 − 1) = 6(x + 1)(x− 1)
the critical points:I f ′(x) = 0 : x = ±1 notice -1 is not in the interval [0,2]II f ′(x) DNE: None3)let’s evaluate f(0), f(1), f(2), then we have
f(0) = 0, f(1) = −4, f(2) = 4
So, on the interval [0,2], f has maximum value 4 at x = 2 and minimum value -4 atx = 1.
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1) cube root can be taken for any x, so the domain of f is all real numbers. f iscontinuous on (−∞,∞), hence it is continuous on [-1,8].2)f ′(x) = 4
3x1/3 − 4
3x−2/3 = 4
3(x1/3 − x−2/3) = 4
3x−2/3(x− 1) the right hand side is done
by factoring out the term that has the lowest power, in this case, x−2/3.Now, this means f ′(x) = 4
3x−13√
x2, so the critical points are
I (f ′(x) = 0)x = 1II (f ′(x)DNE)x = 0
3) let’s evaluate and compare their y values:
f(−1) = 5, f(0) = 0, f(1) = −3, f(8) = 8
So, on the interval [-1,8] f has maximum value 8 at x = 8 and minimum value -3 atx = 1
More examplesa)f(x) = 3− |x− 2|, [1, 4]
1)f is continuous on (−∞,∞), hence it is continuous on [1,4]2)To find f ′(x), we need to simplify f(x) first,
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f ′(2) = limh→0
f(2 + h)− f(2)
h= lim
h→0
3− |(2 + h)− 2| − 3
h
Now notice |2 + h− 2| = |h| so f ′(2) = limh→0
−|h|h
This limit does not exist, so f ′(2) does
not exist. To sum up,I (f ′(x) = 0) : NoneII (f ′(x)DNE)x = 2
3) evaluate and compare y values:
f(1) = 2, f(2) = 3, f(4) = 1
On [1,4], f has maximum value 3 at x = 2 and minimum value 1 at x = 4
b)f(t) = t√
4− t, [−1, 3] (exercise in class)
c)g(s) = 1s−2
, [0, 1]
1)g is continuous on (−∞, 2)∪ (2,∞). and [0, 1] ⊂ (−∞, 2), so g is continuous on [0,1].2)g′(s) = − 1
(s−2)2. This is a rational function. Rational functions reach 0 only when its
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numerator reach 0. critical points:
I (g′(x) = 0) : NoneII (g′(x)DNE)None
3) evaluate and compare:g(0) = −1/2, g(1) = −1
hence, on [0,1], g has maximum value -0.5 at x = 0, and minimum value -1 at x = 1
Core ProblemsP153: 1, 13,17,11,19,33,37,39.
Section 3.6 Max-Min Word Problems
Remark: Apply the theory in 3.5 to find max or min for real life math problems.
STEPS1. Identify the quantity that need to be maximized or minimized.2. Set up a proper function for that quantity.3. Analyze the function to find max and min. (using the method in sect3.5) find thecritical points.4. evaluate and find the max or min5. answer the actual question.
Example 1A farmer is to build a fence using a straight river for one side. What dimension should hebuild the fence if he has 1200 meters of fence and he wants the maximum area possiblein this fence.
sol.1) Need to maximize the area inside the fence.2) Suppose the length of fence is y (along the river), the width is x. Now the relationbetween A, x and y is
A = xy
To find the maximum value for A, we need to look for constraint on x, y. The farmerhas only 1200 meter of fence, that means
2x + y = 1200
Use this we can derive y = 1200 − 2x. Substitute this into the formula for A, we thenhave
A = x(1200− 2x)
Notice x, y are the width and length of the fence, so we must have y > 0, x > 0 hence1200−2x > 0 and x > 0. This gives us 0 < x < 600. Notice that when x = 0 or x = 600,A = 0 according to the formula A = x(1200 − 2x). So it is OK to take [0,600] as thedomain of A = x(1200− 2x).
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dx= (1200− 2x) + x(−2) = 1200− 4x, Hence the critical points are
I (dAdx
= 0)x = 300II (dA
dxDNE)None
c)evaluate and compare: f(0) = 0, f(300) = 300 · 600 = 180000, f(600) = 0. Maximumvalue is 180000 at x = 300. The dimension of the fence should be 300× 600
Example 2A box with square base is to be constructed out of 300in2 of material. What should itsdimension be if it it to have the largest volume.
Sol.1) We hope to maximize the volume V2)suppose the dimension of the box is x× x× y, then the volume is
V = x2y
and the constraint on x, y comes from the limited area 300,
2x2 + 4xy = 300 =⇒ y =300− 2x2
4x
substitute this into the equation for the volume, we have
V = x2(300− 2x2
4x) = x
300− 2x2
4
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Remark Notice the simplification at the last step changes V from a rational functioninto a polynomial function. It is always best to simplify your function in your analysis.Since since x, y are length of the edges of the box, x > 0, y > 0, therefore, x >0, 300−2x2
4x> 0. Hence x should be in (0,
√150). Notice that, if we use V = x300−2x2
4,
it is defined and equal to 0 at x = 0 and x =√
150. The method we use in Sect3.5 isonly valid when we have a closed interval. Since x = 0 and x =
√150 will not produce
a maximum value for V, it is safe to choose maximize the function V = x300−2x2
4, for x
in [0,√
150].3) Let’s apply the max-min theory to find the maximum value of the above function.a)V = x300−2x2
4is a continuous function of x when x is in [0,
√150].
b)dVdx
= 300−2x2
4+ x−4x
4= 300−2x2−4x2
4= 300−6x2
4, Hence the critical points are
I (dVdx
= 0)x =√
50 = 5√
2II (dV
dxDNE)None
c)evaluate and compare: f(0) = 0, f(5√
2) = 5√
2 · 200 = 1000√
2, f(√
150) = 0.Maximum value is 1000
√2 at x = 5
√2. And y = 300−2x2
4x= 200
20√
2= 10√
2= 5
√2. The
dimension of the box should be 5√
2× 5√
2× 5√
2.
Core Problems For this section,it is very important that you do all these problems.Only practice one or two is not enough to get familiar with those STEPS.P164: 11, 5, 20, 25, 9.P164: 13,16,21,27,29,31.P164: 33,45,47.
Webwork:1) A rectangle is inscribed with its base on the x-axis and its upper corners on theparabola y = 11 − x2
5. What are the dimensions of such a rectangle with the greatest
possible area?
2)A car rental agency rents 190 cars per day at a rate of 30 dollars per day. For each 1dollar increase in the daily rate, 5 fewer cars are rented. At what rate should the carsbe rented to produce the maximum income, and what is the maximum income?hint: this is an economic problem. For all economic problems we are assuming linearrelationship between price per units and the number of units. This means y = mx + b,y is the price per unit,x is the number of units. The key is to find m and b.For this problem, let x be the daily price, y be the number of cars rented per day. Whenx=30, y=190. When x=31,y=185. Hence m = −5 and y = −5(x− 30) + 190. The totalincome should be (number of cars rented)× (price per day). So denote the income perday by R, we have R = xy where y = −5(x− 30) + 190. Once you set up the functionfor R. you should be able to maximize it.
3)A commercial cherry grower estimates from past records that if 49 trees are plantedper acre, each tree will yield 27 pounds of cherries for a growing season. Each additionaltree per acre (up to 20) results in a decrease in yield per tree of 1 pound. How manytrees per acre should be planted to maximize yield per acre, and what is the maximumyield?
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hint: this is an economic problem. For all economic problems we are assuming linearrelationship between input and output. This means y = mx+ b, y is the yield per tree,xis the number of trees planted. The key is to find m and b.4)A box is to be made out of a 8 by 16 piece of cardboard. Squares of equal size willbe cut out of each corner, and then the ends and sides will be folded up to form a boxwith an open top. Find the length L, width W , and height H of the resulting box thatmaximizes the volume. (Assume that W ≤ L).Detailed example in the book Page 152 example 2. Let x denote the edge length of thecutout squares. Volume V can be expressed as a polynomial function of x.
5)A parcel delivery service will deliver a package only if the length plus the girth (dis-tance around) does not exceed 104 inches. Find the maximum volume of a rectangularbox with square ends that satisfies the delivery company’s requirements.
6)A fence is to be built to enclose a rectangular area of 300 square feet. The fence alongthree sides is to be made of material that costs 6 dollars per foot, and the material forthe fourth side costs 16 dollars per foot. Find the length L and width W (with W ≤ L)of the enclosure that is most economical to construct. (not required)
7)A small resort is situated on an island. The closest point on the mainland is a pointP which is exactly 3 miles from the resort. The closest source of fresh water is 10 milesdown the shoreline from P . The resort is planning to lay pipe to bring water from thatsource. It will run along the shore for some distance and then turn and cross the waterto the resort.
However, running pipe underwater costs 2 times as much as running pipe down theshore. At what distance down the shore from P should the pipe turn and head for theresort?
Similar to the ”powerline through the park” project in 2200lab classes.
8)The manager of a large apartment complex knows from experience that 80 units willbe occupied if the rent is 324 dollars per month. A market survey suggests that, onthe average, one additional unit will remain vacant for each 2 dollar increase in rent.Similarly, one additional unit will be occupied for each 2 dollar decrease in rent. Whatrent should the manager charge to maximize revenue?hint: this is an economic problem. For all economic problems we are assuming linear
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relationship between money and units. This means y = mx+ b, y is the number of unitsrented,x is the price per month. The key is to find m and b.
9)Let Q = (0, 3) and R = (8, 4) be given points in the plane. We want to find the pointP = (x, 0) on the x-axis such that the sum of distances PQ+PR is as small as possible.(Before proceeding with this problem, draw a picture!)
10)Centerville is the headquarters of Greedy Cablevision Inc. The cable company isabout to expand service to two nearby towns, Springfield and Shelbyville. There needsto be cable connecting Centerville to both towns. The idea is to save on the cost of cableby arranging the cable in a Y-shaped configuation. Centerville is located at (9, 0) inthe xy-plane, Springfield is at (0, 6), and Shelbyville is at (0,−6). The cable runs fromCenterville to some point (x, 0) on the x-axis where it splits into two branches goingto Springfield and Shelbyville. Find the location (x, 0) that will minimize the amountof cable between the 3 towns and compute the amount of cable needed. Justify youranswer.
Remark: Before we look at more examples for the Max-Min applications, let us practicea little more on finding derivatives.
Let’s start by looking at some of problems in webwork:
Set 3.2:problem 7If f(x) = 4x
√x + 2
x2√
x, find f ′(x).
Set 3.3:problem 3Let f(x) = (x3 + 3x + 8)4 then f ′(x) =?
Set 3.3:problem 4If f(x) = (5x + 8)−1, find f ′(x).
Set 3.3:problem 5Let f(x) =
√2x2 + 2x + 7, find f ′(x)
Set 3.3:problem 6
Suppose that y = −4√5x2+5
. Finddy
dx.
Set 3.3:problem 8Let f(x) = (9x2 + 8)3(6x2 − 7)12
Set 3.6:problem 8The manager of a large apartment complex knows from experience that 80 units willbe occupied if the rent is 324 dollars per month. A market survey suggests that, onthe average, one additional unit will remain vacant for each 2 dollar increase in rent.Similarly, one additional unit will be occupied for each 2 dollar decrease in rent. Whatrent should the manager charge to maximize revenue?
Set 3.6b:problem 5Find the point on the line 5x + 7y + 3 = 0 which is closest to the point ( 4, 2 ).
Section 3.7 The Derivative of Trigonometric Functions
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Recall basic trig limit
limx→0
sin(x)
x= 1
limx→0
cos(x)− 1
x= 0
Trig formulas for the sum of two angles
sin(x + y) = sin(x) cos(y) + sin(y) cos(x)cos(x + y) = cos(x) cos(y)− sin(y) sin(x)
Theorem 1 Derivatives of Sines and Cosines
The functions f(x) = sin(x) and g(x) = cos(x) are differentiable for all x, and
d
dx(sin(x)) = cos(x),
d
dx(cos(x)) = − sin(x)
Proof.To differentiate f(x) = sin(x), we begin with the definition of the derivative,
f ′(x) = limh→0
f(x + h)− f(x)
h
= limh→0
sin(x + h)− sin(x)
h
( apply formula above) = limh→0
(sin x cos h + sin h cos x)− sin(x)
h
= limh→0
[sin x(
cos h− 1
h) + cos x
sin h
h
]= sin x
(limh→0
(cos h− 1
h)
)+ cos x lim
h→0
sin h
h(use basic trig limit) = sin x(0) + cos x(1) = cos x
To differentiate f(x) = cos(x), we begin with the definition of the derivative,
f ′(x) = limh→0
f(x + h)− f(x)
h
= limh→0
cos(x + h)− cos(x)
h
( apply formula above) = limh→0
(cos x cos h− sin h sin x)− cos(x)
h
= limh→0
[cos x(
cos h− 1
h)− sin x
sin h
h
]= cos x
(limh→0
(cos h− 1
h)
)− sin x lim
h→0
sin h
h(use basic trig limit) = cos x(0)− sin x(1) = − sin x
Example 1 Calculated
dx(cos3 x)
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f (x) sin x cos x tan x cot x sec x csc xddxf (x) cos x − sin x sec2 x − csc2 x sec x tan x − csc x cot x
Table 1: derivative for trig functions
Sol.This is a composition of functions. suppose we denoteh(x) = cos3 x, f(x) = x3, g(x) = cos x, then h(x) = f(g(x)). Notice f ′(x) = 3x2, g′(x) =− sin x so
d
dx(cos3 x) = d
dxf(g(x))
= f ′(g(x))g′(x)= 3(cos x)2(− sin x)= −3 cos2 x sin x
Theorem 2 The Derivatives of remaining Trig functions
d
dx(tan(x)) = sec2(x),
d
dx(cot(x)) = − csc2(x),
d
dx(sec(x)) = sec(x) tan(x),
d
dx(csc(x)) = − csc(x) cot(x)
Proof.
d
dx(tan x) =
d
dx(sin x
cos x)
=cos x(sin x)′ − sin x(cos x)′
cos2 x
=cos x cos x− sin x(− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x
=1
cos2 x= sec2(x),
d
dx(csc x) =
d
dx(
1
sin x) =
−(sin x)′
sin2 x
=− cos x
sin2 x=− cos x
sin x
1
sin x
= − cot x1
sin x= − cot x csc x,
RemarkThere are certain patterns in these formulas that will be helpful in memorizing them. Itwill also be beneficial to write them in a little table. Note that the derivative formulasfor the three cofunctions are those involving minus signs.
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The chain rule version of trignometric differentiation formulas
d
dx(sin u) = cos u
du
dxd
dx(cos u) = − sin u
du
dxd
dx(tan u) = sec2 u
du
dxd
dx(cot u) = − csc2 u
du
dxd
dx(sec u) = sec u tan u
du
dxd
dx(csc u) = − csc u cot u
du
dx
Examples
(1). If f(x) =2x2 tan x
sec x, find f ′(x). (webwork set 3.7 problem 6)
Sol.First, let’s simplify if possible,
f(x) =2x2 tan x
sec x= 2x2 tan x
1
sec x= 2x2 sin x
cos xcos x = 2x2 sin x
therefore,f ′(x) = 2x2(sin x)′ + sin x(2x2)′ = 2x2 cos x + 4x sin x
(2). Findd
dxsin5(x7 + 1). (webwork set 3.7 problem 12)
sol.This is a composition of functions. We know that the derivative of the compositionfunction of two functions f(g(x)) can be calculated by f ′(g(x))g′(x). If the compositionhas three ”layers”, say, h(f(g(x))), then its derivative is h′(f(g(x)))f ′(g(x))g′(x).
x → x7 + 1 → sin(x7 + 1) → sin5(x7 + 1)
Let h(x) = x5, f(x) = sin x, g(x) = x7 + 1, then sin5(x7 + 1) = h(f(g(x))). so
d
dxsin5(x7 + 1) = h′(f(g(x)))f ′(g(x))g′(x) = 5 sin4(x7 + 1) cos(x7 + 1)7x6
The following two problems is for class practice.
(3). Findd
dx(cos x2 − tan x3).
(4). Findd
dx(x2 sec x3).
(5) Find the maximum value of y = 3 sin x cos x on [0, π].
Sol.a) y = 3 sin x cos x is continuous on (−∞,∞), hence it is continuous on [0, π].
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b)dydx
= 3(sin x(cos x)′ + cos x(sin x)′)= 3(sin x(− sin x) + cos x cos x)= 3(cos2 x− sin2 x)
Hence the critical points in [0, π] is when cos2 x− sin2 x = 0. By double angle formula
cos 2x = 0. This only happens when 2x = π2
or 3π2
. (notice when x is in [0, π], 2x is in[0, 2π])
therefore, the critical points:
(I dydx
= 0)x = π4
or 3π4
(II dydx
DNE ) None
c) Let’s evaluate the y value at the end points and critical points:
f(0) = 0, f(π
4) =
3
2, f(
3π
4) = −3
2, f(π) = 0
Hence, max value 32
at x = π4.
Core Problems
P177: 1,3,5,9,11,13,67,72.P177: 15,41,43,51,59,73,75,77,79.
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Answers for 3) and 4)
3)d
dx(cos x2 − tan x3) = − sin x2(2x)− sec2 x3(3x2).
4).d
dx(x2 sec x3) = x2 sec x3 tan x33x2 + sec x3(2x).
Section 3.8 Exponential and Logarithmic Functions
Review
A exponential function is a function of the form
f(x) = ax
where a > 0.
The laws of exponents:
ar+s = ar · as, (ar)s = ar·s
ar =1
ar, (ab)r = ar · br
A base a logarithm function loga x is defined as following:loga x is the power to which a must be raised to get x. Thus
y = loga x if and only if ay = x
The number e is given by the limit
e = limn→∞
(1 +
1
n
)n
The base e logarithm function is called natural logarithm function and is commonlydenoted by the special symbol ln:
ln x = loge x(x > 0)
The laws of logarithms
ln(ab) = ln a + ln b, ln(a
b) = ln a− ln b,
ln ab = b ln a, ln 1 = 0
A graph of the natural exponential function y = ex and natural logarithm functiony = ln x is given below.
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d
dxex = ex.
Therefore, by Chain Rule, we have
d
dxeu = eu du
dx.
Example 1 Findd
dxe(x2+5x).
(webwork set 3.8 problem 5)
Sol.Let u = x2 + 5x, then
d
dxe(x2+5x) =
d
dxeu.
By Theorem 1, we have
d
dxeu = eu du
dx= e(x2+5x)d(x2 + 5x)
dx.
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Therefore,d
dxe(x2+5x) = e(x2+5x)(2x + 5)
Example 2 Findd
dx
(x2e4 cos x
).
(webwork set 3.8 problem 4)
Sol.Apply Product Rule, we see
d
dx
(x2e4 cos x
)= x2 d
dxe4 cos x + e4 cos x d
dxx2.
We know that ddx
x2 = 2x. So the key is to find ddx
e4 cos x. Let u = 4 cos x, then by Theorem1
d
dxe4 cos x =
d
dxeu = eu du
dx
Therefore,d
dxe4 cos x = e4 cos x d
dx4 cos x = e4 cos x4(− sin x)
Finally,d
dx
(x2e4 cos x
)= x2
(e4 cos x4(− sin x)
)+ 2xe4 cos x.
Or,d
dx
(x2e4 cos x
)= −4x2 sin xe4 cos x + 2xe4 cos x.
Theorem 2 If f(x) = ln x, then f ′(x) = 1x. i.e.
d
dx(ln x) =
1
x.
Therefore, by Chain Rule, we have
d
dx(ln u) =
1
u
du
dx.
Remark If h(x) = ln u, and u = g(x), then
h′(x) =d
dx(ln u) =
1
u
du
dx=
1
g(x)g′(x)
Proof of ”If f(x) = ln x, then f ′(x) = 1x.”
Let y = ln x, then by the definition of logarithmic functions we know
x = ey
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Calculate derivative functions of both sides, we will find that the derivative of the leftside is
d
dxx = 1
Since y is a function of x, then right side is a composition function. By Chain Rule inTheorem 1 we see its derivative is
d
dxey = ey dy
dx
Since x = ey, their derivatives should be equal, hence
1 = ey dy
dx
This givesdy
dx=
1
ey=
1
x
Remark The technique in this proof, taking derivatives of both sides of an equation,will be helpful later in the understanding of implicit differentiations.
Example 3d
dx(ln(x2 + 8))
(webwork set 3.8 problem 8)
Sol.Since this is a composition function with ln as the ”outer function”, by the earlierremark, we know its derivative should be
d
dx(ln(x2 + 8)) =
ddx
(x2 + 8)
x2 + 8=
2x
x2 + 8
Example 4d
dx
√1 + ln(x2 + 2)
(webwork set 3.8 problem 11)
Sol.Let u = 1 + ln(x2 + 2), then we are calculating:
d
dx
√1 + ln(x2 + 2) =
d
dx
√u =
d
du
√udu
dx=
1
2√
u
du
dx
This means
d
dx
√1 + ln(x2 + 2) =
1
2√
1 + ln(x2 + 2)
d
dx(1 + ln(x2 + 2))
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Nowd
dx(1 + ln(x2 + 2)) =
d
dx(ln(x2 + 2)) =
(x2 + 2)′
x2 + 2=
2x
x2 + 2
Henced
dx
√1 + ln(x2 + 2) =
1
2√
1 + ln(x2 + 2)
2x
(x2 + 2)
Example 5d
dxln
[(x + 5)3
(x + 1)(x2 + 1)5/2
](webwork set 3.8 problem 14)
Sol.This is also a compostional function with ln as the ”outer function”. But the expres-sion insider the ln calculation seems to be complicated. The key is to simplify beforedifferentiation.
ln
[(x + 5)3
(x + 1)(x2 + 1)5/2
]= 3 ln(x + 5)− ln(x + 1)− 5
2ln(x2 + 1)
This means
d
dxln
[(x + 5)3
(x + 1)(x2 + 1)5/2
]= 3
1
x + 5− 1
x + 1− 5
2
2x
(x2 + 1)
Remark Now we know that by using Chain Rule we can differentiate functions likeeg(x). But how do we differentiate functions that are in the form f(x)g(x) ? Here is auseful formula
ab = eln ab
= eb ln a.
Using this, we can transform any exponent formula into the base e.
f(x)g(x) = eln f(x)g(x)
= eg(x) ln f(x)
then we can apply chain rule again.
Example 6 Findd
dxxx2
.
(webwork set 3.8 problem 14)
Sol.By the above remark,
d
dxxx2
=d
dxex2 ln x.
This is a compostion function with natural exponential function as the ”outer function”.By the chain rule in Theorem 1, we have
d
dxex2 ln x = ex2 ln x d
dx(x2 ln x).
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Use product rule to get
d
dxx2 ln x = x2(ln x)′ + (x2)′ ln x = x2 1
x+ 2x ln x = x + 2x ln x.
Henced
dxxx2
= ex2 ln x(x + 2x ln x).
Core Problems
P192: 1,3,5,9,17,19,23,33,37,59.
Section 3.9 Implicit Differentiation and Related Rates
Most of the functions we have so far have been defined explicitly, such as y = f(x), y =x7, y = sin x, y = ex. Notice y is expressed explicitly in terms of x. However, a functioncan also be defined ’implicitly’ by an equation that can be solved for y in terms of x.
Example 1
Given a circle, x2 + y2 = 2 finddy
dxat (1, 1).
Sol.(I) Assume y is a differentiable function of x at (1,1).
(II) Differentiate on both sides of x2 + y2 = 2 , we have
d
dx(x2 + y2) =
d
dx(2)
2x + 2ydy
dx= 0
notice that in the second line, we obtainedd
dx(y2) = 2y
dy
dxby using chain Rule.
(III) Solve 2x + 2ydy
dx= 0 for
dy
dxwe have
dy
dx= −x
y
(IV) Evaluate atdy
dxat (1,1)
dy
dx|x=1,y=1= −1
Remark
The technique above is called Implicit Differentiation. It is useful when we do not havean explicit formula for a function, but we are able to obtain an equation that is relatedto the desired function. As we can see, in this example, equation x2 + y2 = 2 did not
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give us explicit formula for y. And because for every x value there will be two possibley value that will be valid solutions for this equation. This means this equation is nota function. To give a explicit formula for y we will have to give individual solution fory > 0 and y < 0.
y =
{ √2− x2, y ≥ 0
−√
2− x2, y < 0
Obviously, at (1,1), y > 0 so y =√
2− x2. Hence
dy
dx=
1
2√
2− x2(−2x) =
−x√2− x2
Evaluate at x = 1, we will again have dydx
= −1
Although in this particular example, we can use both method to find the right answer,there are situations where explicit formulas are too difficult to obtain. So, ImplicitDifferentiation is a valuable tool when it comes to finding derivatives or calculating rateof change.
Given E(x, y) = c To finddy
dxat specified point.
STEPS
(II) Assume y is a differentiable function of x.
(II) Differentiate on both sides of E(x, y) = c,
d
dxE(x, y) =
d
dx(c)
(III) solved
dxE(x, y) =
d
dx(c) for
dy
dx.
(VI) Evaluate at the designated point.
Example 2
Given x3 + y3 = 3xy finddy
dx.
Sol.(I) Assume y is a differentiable function of x.
(II) Differentiate on both sides of x3 + y3 = 3xy , we have
d
dx(x3 + y3) =
d
dx(3xy)
3x2 + 3y2 dy
dx= 3(x
dy
dx+ y)
notice that in the second line, we obtainedd
dx(y3) = 3y2 dy
dxby using chain Rule and we
obtainedd
dx(xy) = x
dy
dx+ y by using procduct rule.
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(III) Solve 3x2 + 3y2 dy
dx= 3(x
dy
dx+ y) for
dy
dxwe have
x2 + y2 dy
dx= x
dy
dx+ y
y2 dy
dx− x
dy
dx= y − x2
dy
dx=
y − x2
y2 − x
More Examples
(1) For the equation given below, evaluate y′ at the point (2,1).5x3 − 2y = ln(y) + 38− ln(1).
(2) For the equation given below, evaluate y′ at the point (2,1).2exy − 3x = y + 7.78.
Related Rates
A related-rates problem involves two or more quantities that vary with time and anequation that expresses some relationship between these quantities. Typically, the valuesof these quantities at some instant are given, together with all their time rates of changebut one. The problem is usually to find the time rate of change that is NOT given, atsome instant specified in the problem. Implicit differentiation, with respect to t, of theequation that relates the given quantities will produce an equation that relates the ratesof change of the given quantities. This is the key to solving a related-rates problem.
Example 3
A rocket that is launched vertically is tracked by a radar station located on the ground3 mi from the launch site. What is the vertical speed of the rocket at the instant thatits distance from the radar station is 5 mi and this distance is increasing at the rate of5000 mi/h?
Sol.(I) Draw a graph and label the changing quantities as variables.
We denote the altitude of the rocket (in miles) by y and denote its distance from theradar by z. Notice that they are both functions of time t.
We are trying to decide the vertical speed of the rocket, whick equals the velocity. Thevelocity is the derivative of the position function, hence the key is to find dy
dt.
(II) Record the values of variables or rates of change, as given in the problems.
We are givendz
dt= 5000 when z = 5.
(III) Use the graph to determine an equation that relates these variables.
We apply Pythagorean theorem to the right triangle in the figure and obtain
y2 + 9 = z2
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From this, we see that when z = 5, y = 4.
(VI) Implicit differentiate y2 + 9 = z2 with respect to t, we then have
2ydy
dt= 2z
dz
dt
(V) Substitute the given numerical data in the resulting equation, and then solve forthe unknown.
We know z = 5, y = 4, anddz
dt= 5000, so
2(4)dy
dt= 2(5)(5000).
Solve this, then we find dydt
= 6250.
Example 4
A street light is at the top of a 14 ft tall pole. A woman 6 ft tall walks away from thepole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadowmoving when she is 35 ft from the base of the pole? (webwork set 3.9 problem 17)
Note: You should draw a picture of a right triangle with the vertical side representing thepole, and the other end of the hypotenuse representing the tip of the woman’s shadow.Where does the woman fit into this picture? Label her position as a variable, and labelthe tip of her shadow as another variable. You might like to use similar triangles to finda relationship between these two variables.
Sol.(I) Draw a graph and label the changing quantities as variables.
Denote the distance from the base of the pole to the woman’s feet as x. Denote thedistance from the base of the pole to the woman’s shadow as z. As time t changes, bothx and z are changing. So they are both functions of time t.
(II) use the graph to determine an equation that relates these variables.
We see two right triangles in this graph. Using similar triangle property, we know
z
14=
z − x
6
(III) Record the values of variables or rates of change, as given in the problems.
The woman’s walking speed is the velocity, i.e. the derivative of the position function
x. Hence we getdx
dt= 7.
The question is to figure out how fast is the tip of her shadow moving when she is 35 ftfrom the base of the pole. This is asking dz
dtwhen x = 35.
To sum it up, we are givenz
14=
z − x
6and dx
dt= 7. We need to calculate
dz
dtwhen
x = 35.
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(VI) Differentiatez
14=
z − x
6implicitly with respect to t.
d
dt
z
14=
d
dt
z − x
6114
dz
dt= 1
6(dz
dt− dx
dt)
6dz
dt= 14(
dz
dt− dx
dt)
(V) Substitute the given numerical data in the resulting equation, and then solve forthe unknown.
Sincedx
dt= 7 we can solve
6dz
dt= 14(
dz
dt− 7)
and determine thatdz
dt= 98/8 = 12.25
More examples
(1)(webwork set 3.9 problem 11) A rocket is launched vertically upward from a point8 miles west of an observer on the ground. What is the speed of the rocket when theangle of elevation (from the horizontal) of the observer’s line of sight to the rocket is53 degrees and is increasing at 1 degrees per second? (First give it in miles per second,and then in miles per hour)
Note: the easiest mistake in the problem is to forget to convert degree into radian.
(I) Draw a graph. Label the altitude of the rocket as y, and label the angle of elevationas θ.
(II) Given information: when θ = 53, dθdt
= 1 degree per seconds, i.e. dθdt
= π180
radianper second.
(III) according to the graph:
tan(θ) =y
8
(VI) Implicit Differentiation gives:
d
dt[tan(θ)] =
d
dt[y
8]
notice both θ and y are functions of time t
sec2 θdθ
dt=
1
8
dy
dt
(V) solve for dydt
, we have
dy
dt= 8 sec2 θ
dθ
dt= 8 sec2(53o) · π
180= 8 sec2(53
π
180) · π
180= 0.385515
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This is miles/sec. To change it into Mile/hour, you need to multiply your answer by3600.
(2)Helium is pumped into a spherical balloon at a rate of 2 cubic feet per second. Howfast is the radius increasing after 2 minutes? (webwork set 3.9 problem 12) Note: Thevolume of a sphere is given by V = 4
3πr3.
Hint: To find the radius of the balloon at 2 minutes, assume that it is empty (has volume0) at time t=0 and use your knowledge of dV
dt(i.e., that the volume is increasing at a
constant rate) to find V at 2 minutes.
Given information: dVdt
= 2, V = 43πr3
Need to find: drdt
Take V = 43πr3. and differentiate with respect to time t on both sides.
dV
dt=
4
3π3r2dr
dt
As in the hint, we assume when t = 0, r = 0, V = 0, and try to find r when t = 2
If dVdt
= 2, for all t then V is a linear function of time t : V = mt + b and m = 2 cubicfeet per second. When t = 0, V = 0, so b = 0. We then have V = 2t. When t = 2minute,that is t = 120 second. V = 240.
Because V = 43πr3. when V = 240, we find r = (180
π)1/3
Now we can solve dVdt
= 4πr2 drdt
for drdt
:
2 = 4π(180
π)2/3dr
dt
dr
dt= (2π(
180
π)2/3)−1 = 0.0107087
Core Problems
Implicit Differentiation, P200: 7,11,13,19,23,25,31
Related Rates(day 1), P202:45,55,39,43,51
Related Rates(day 2), P202:60,38,61,56,68,47.
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