Chapter 3
3.1
Quick Review
20 2
4 0
2
In Exercises 1 4, evaluate the indicated limit algebraically
2 4 3
1. lim 2. lim
2 2 8
3. lim 4. lim
2
5. Find the slope of the line tangent to the parabola 1 at its vertex.
6. By consider
h x x y h x h y x y x y x
3 2ing the graph of 3 2,find the intervals on which is increasing.
f x x x f
Quick Review
2 1 1 0 1In Exercises 7 10, let
2, 1
1 , 1 7. Find lim and lim .
8. Find lim 1 .
9. Does lim exist? Explain. 10. Is continuous? Explain.
x x h x x x f x x x
f x f x
Quick Review Solutions
20 2
4 0
2
5
In Exercises 1 4, evaluate the indicated limit algebraically
2 4 3
1. lim 2. lim
2 2 8
3. lim 4. lim
2 5. Find the slope of the line tang
4
e
2
nt to the parabola 1 at its vertex.
6. y
0
B c
1 8 h x x y h x h y x y x y x
3 2onsidering the graph of 3 2,find the intervals on which is increasing. ( ,0] and [2, )
f x x x f
Quick Review Solutions
2 1 1 0 1In Exercises 7 10, let
2, 1
1 , 1 7. Find lim and lim .
8. Find lim 1 . 9.
0;3
0
No. The
Does lim exist? Ex
two one-sided limits are different.
plain.
10. Is continuous? Expl
x x h x x x f x x x
f x f x
f h f x f
No, is discontinuous at 1
What you’ll learn about
Definition of a Derivative
Notation
Relationship between the Graphs of f and f '
Graphing the Derivative from Data
One-sided Derivatives
… and why
Definition of Derivative
0
The of the function with respect to the variable is the function whose value at is
lim provided the limit exists.
h
f x
f x x
f x h f x f x
h
Differentiable Function
The domain of , the set of points in the domain of for which the limit exists, may be smaller than the domain of . If exists, we say that
at . A
f f
f f x f
x
Example
Definition of Derivative
2Differentiate f x x
0 2 2 substitute 02 2 2
2
expanded 0
2
Applying the definition, we have lim lim 2 lim 2 lim h h x h h
f x h f x f x
h x h x
h
x xh h x h
x h h
Derivative at a Point (alternate)
The derivative of the function at the point where is the limit lim
provided the limit exists.
x a
f x a
f x f a f a
x a
Notation
There are many ways to denote the derivative of a function ( ). Besides '( ), the most common notations are:
" prime" Nice and brief, but does not name
y f x f x
y y
the independent variable. " " or "the derivative Names both variables and
of with respect to " uses for derivative. " " or "the d
dy
dy dx
dx y x d
df
df dx dx
erivative Emphasizes the function's name. of with respect to "f x
Relationships between the
Graphs of f and f’
Because we can think of the derivative at a point in graphical
terms as slope, we can get a good idea of what the graph of the
function
f’
looks like by
estimating the slopes
at various points
along the graph of
f
.
Graphing the Derivative from Data
Discrete points plotted from sets of data do not yield a continuous
curve, but we have seen that the shape and pattern of the graphed
points (called a scatter plot) can be meaningful nonetheless. It is
often possible to fit a curve to the points using regression
One-sided Derivatives
0 0A function is if
it has a derivative at every interior point on the interval, and if the limits
lim the
lim
h
h
y f x
f a h f a h
f b h f b
differentiable on a closed interval a, b
right - hand derivative at a
the
exist at the endpoints. In the right-hand derivative, is positive and approaches a from the right. In the left-hand derivative, h is negative and approaches
h
h a h
b h
left - hand derivative at b
from the left.
One-sided Derivatives
Right-hand and left-hand derivatives may be defined at any point
of a function’s domain.
The usual relationship between one-sided and two-sided limits
holds for derivatives. Theorem 3, Section 2.1, allows us to
Example
One-sided Derivatives
Show that the following function has left-hand and right-hand derivatives at 0, but no derivative there.
, 0 , 0 x x x y x x
0 0 0 0Left-hand derivative: Right-hand derivative:
0 0 0 0
lim lim
lim 1 lim 1
The derivatives are not equal at 0. The function does not
3.2
Quick Review
0 0
0
In Exercises 1-5, tell whether the limit could be used to define assuming that is differentiable at .
1. lim 2. lim
3. lim 4. lim
5. lim
h h
x a x a
h
f a
f a
f a h f a f a h f h
h h
f x f a f a f x
x a a x
f a h f a h h
Quick Review
4 3 3 4
6. Find the domain of the function . 7. Find the domain of the function .
8. Find the range of the function 2 3.
9. Find the slope of the line 5 3.2 .
3 0.001 3 0.001 10. If 5 , find
0
y x y x
y x
y x
f f
f x x
Quick Review Solutions
0 0 0In Exercises 1-5, tell whether the limit could be used to define assuming that is differentiable at .
1. lim 2. lim
3. lim
Yes No
Yes 4. lim Ye
5. l
s
im
h h
x a x a
h
f a
f a
f a h f a f a h f h
h h
f x f a f a f x
x a a x
f a h f a
Quick Review Solutions
4 3 3 46. Find the domain of the function . 7. Find the domain of the function .
8. Find the range of the function 2 3.
9. Find the slope of the line 5 3.2 .
1
All Reals
[0, )
[3, )
3.
0. If 5 , f
2 i y x y x y x y x
f x x
3 0.001
3 0.001
nd
0.002 5
What you’ll learn about
How
f’(a)
Might Fail to Exist
Differentiability Implies Local Linearity
Derivatives on a Calculator
Differentiability Implies Continuity
Intermediate Value Theorem for Derivatives
… and why
How f’(a) Might Fail to Exist
A function will not have a derivative at a point , where the slopes of the secant lines,
fail to approach a limit as approaches .
The next figures illustrate four different instances w
P a f a f x f a
x a
x a
How f’(a) Might Fail to Exist
1. a corner, where the one-sided derivatives differ;
How f’(a) Might Fail to Exist
232. a cusp, where the slopes of the secant lines approach from one side and approach - from the other (an extreme case of a corner);
f x x
How f’(a) Might Fail to Exist
33. A vertical tangent, where the slopes of the secant lines approach either or - from both sides;
f x x
How f’(a) Might Fail to Exist
4. a discontinuity (which will cause one or both of the one-sided derivatives to be nonexistent).
1, 0
1, 0
x U x
x
Example
How
f’(a)
Might Fail to Exist
3Show that the function is not differentiable at 0.
, 0
4 , 0
x x x
f x
x x
The right-hand derivative is 4. The left-hand derivative is 0.
How f’(a) Might Fail to Exist
Most of the functions we encounter in calculus are differentiable
wherever they are defined, which means they will
not
have
corners, cusps, vertical tangent lines or points of discontinuity
Differentiability Implies Local Linearity
A good way to think of differentiable functions is that they are
locally linear
; that is, a function that is differentiable at
a
closely resembles its own tangent line very close to
a.
In the jargon of graphing calculators, differentiable curves will
“straighten out” when we zoom in on them at a point of
Derivatives on a Calculator
Many graphing utilities can approximate derivatives numerically with good accuracy at most points of their domains. For small values of ,
the difference quotient
is often a good numerical app
h
f a h f a h
roximation of .
However, the same value of will usually yield a better approximation if we use the symmetric difference quotient
2
which is what our graphing calculator uses to calculate N
f a h
f a h f a h h
DER , the .
The as a function is denoted by NDER .
f a
f a
f f x
Example
Derivatives on a Calculator
2Find the numerical derivative of the function 3 at the point 2. Use a calculator with 0.001.
f x x
x h
Derivatives on a Calculator
Because of the method used internally by the calculator, you will
sometimes get a derivative value at a nondifferentiable point.
Differentiability Implies Continuity
If has a derivative at f x a , then is continuous at f x a .
Intermediate Value Theorem for
Derivatives
Not every function can be a derivative.
If and are any two points in an interval on which is differentiable, then takes on every value between
and .
a b f
f f a f b
3.3
Quick Review
1 2 1 2 4 3 2 2 2 1 2 1 2 3In Exercises 1-6, write the expression as a sum of powers of .
1. 2 1 2.
1
2 5 3 2 4
3. 3 4.
2
5. 2 1 6.
x x x x x x x x
x x x
Quick Review
3 2
6
7. Find the positive roots of the equation 2 5 2 6 0
and evaluate the function 500 at each root. Round your answers to the nearest integer, but only in the final step.
8. If 7 for all real
x x x
y x
f x
numbers , find
a 10 b 0
c d lim
x
f f
f x f a
Quick Review
2 15
9. Find the derivatives of these functions with respect to .
a b c
10. Find the derivatives of these functions with respect to using the definition of the derivative.
a b
x
f x f x f x
x
x
f x f x
x
Quick Review Solutions
1 2 1 2 4 32 1 1
2 1 2 2 2
2 2 2 1 2 1 2 3 2 2 3
3 2 5 2
2
In Exercises 1-6, write the expression as a sum of powers of .
1. 2 1 2.
1
2 5 3 2 4
3. 3 4.
2
5. 2 1 6.
x x x x x
x x x x x x
x x x x x x x x
x x x
Quick Review Solutions
3
6
2 6
7. Find the positive roots of the equation 2 5 2 6 0
and evaluate the function 500 at each root. Round your answers to the nearest integer, but only in the final s
At 1.173, 5
t
0
ep
0 1305 A
.
t
x x
x x x
y x x
68. If 7 for all real numbers , fin
2.394, 500 94,212
7
d
a 10 b 0 7
7 d i 0
c l m
x a
f x x
f f
f x f a
Quick Review Solutions
2 15
9. Find the derivatives of these functions with respect to .
a b c
10. Find the derivatives of these functions with respect to using the definition of the derivative.
a
0 0 0
1
b
x
f x f x f x
x
x
f x f x
x
f x
What you’ll learn about
Positive Integer Powers, Multiples, Sums
and Differences
Products and Quotients
Negative Integer Powers of
x
Second and Higher Order Derivatives
… and why
Rule 1 Derivative of a Constant Function
If is the function with the constant value , then 0
This means that the derivative of every constant function is the zero function.
f c
Rule 2 Power Rule for Positive Integer
Powers of x.
1If is a positive integer, then
The Power Rule says:
To differentiate , multiply by and subtract 1 from the exponent.
n n
n
n
d
x nx dx
x n
Rule 3 The Constant Multiple Rule
If is a differentiable function of and is a constant, then
This says that if a differentiable function is multiplied by a constant, then its derivative is multiplied by the same cons
u x c
d du
cu c dx dx
Rule 4 The Sum and Difference Rule
If and are differentiable functions of , then their sum and differences
are differentiable at every point where and are differentiable. At such points, .
u v x
u v
d du dv
u v
Example
Positive Integer Powers,
Multiples, Sums, and Differences
4 2 3
Differentiate the polynomial 2 19 4
That is, find .
y x x x
dy dx
4
2
Sum and Difference Rule 3
Constant and Power Rules
By Rule 4 we can differentiate the polynomial term-by-term, applying Rules 1 through 3.
3
2 19
4 3
4 2 2 0
4
dy d d d d
x x x
dx dx dx dx dx
x x
Example
Positive Integer Powers,
Multiples, Sums, and Differences
4 2
Does the curve 8 2 have any horizontal tangents? If so, where do they occur?
Verify you result by graphing the function.
y x x
4 2
3If any horizontal tangents exist, they will occur where the slope
is equal to zero. To find these points we will set 0 and solve for .
Calculate 8 2 4 16
Set 0 and solve
dy dx dy x dx dy d
x x x x
dx dx dy dx
3 2 2 for 4 16 04 4 0; 4 0 4 0
x
x x
x x x x
Rule 5 The Product Rule
The product of two differentiable functions and is differentiable, and
The derivative of a product is actually the sum of two products.
u v
d dv du
Example
Using the Product Rule
3
2
Find f x if f x x 4 x 3
3 2
3 2 3 2 2
4 4 2
4 2
Using the Product Rule with 4 and 3,gives
4 3 4 2 3 3
2 8 3 9
5 9 8
u x v x
d
f x x x x x x x
dx
x x x x
x x x
Rule 6 The Quotient Rule
2
At a point where 0, the quotient of two differentiable functions is differentiable, and
Since order is important in subtraction, be sure to set up the numerator of the
u
v y
v
du dv
v u
d u dx dx
dx v v
Example
Using the Quotient Rule
32 4Find if
3
x f x f x
x
3 23 2 2 3
2
2 2
4 2 4
2 2
4 2
2 2
Using the Quotient Rule with 4 and 3,gives
4 3 3 4 2
3 3
3 9 2 8
3
9 8
3
u x v x
x x x x x
d f x
dx x x
x x x x
x
x x x
Rule 7 Power Rule for Negative Integer
Powers of x
1If is a negative integer and 0, then .
This is basically the same as Rule 2 except now is negative.
n n
n x
d
x nx dx
n
Example
Negative Integer Powers of
x
1
Find an equation for the line tangent to the curve y at the point 1,1 .
x
1 2 2Rewrite the function as and use the Power Rule to find the derivative.
1
1
1
Evaluate 1 = 1 1
The line through 1,1 with slope 1 is
1 1 1
2
This shows the graph of the funct
y x y x x y m y x y x
ion and its tangent line at (1, 1). 1
y x
2
Second and Higher Order Derivatives
2 2
The derivative is called the of with respect to . The first derivative may itself be a differentiable function of . If so,
its derivative, ,
dy
y first derivative y x
dx
x dy d dy d y
y
dx dx dx dx
3 3
is called the of with respect to . If double prime is differentiable, its derivative,
,
is called the of with respect to .
second derivative y x y y
dy d y y
dx dx
third derivative y x
Second and Higher Order Derivatives
1
The multiple-prime notation begins to lose its usefulness after three primes. So we use " super "
to denote the th derivative of with respect to . Do not confuse the notation with th
n n
n
d
y y y n
dx
n y x
y
Quick Quiz Sections 3.1 – 3.3
You may use a graphing calculator to solve the following problems.
1. Let 1 . Which of the following statements about are true? I. is continuous at 1.
II. is differentiable at 1. III. has
f x x f
f x
f x
f
a corner at 1. A I only
B II only C III only
D I and III only
Quick Quiz Sections 3.1 – 3.3
You may use a graphing calculator to solve the following problems.
1. Let 1 . Which of the following statements about are true? I. is continuous at 1.
II. is differentiable at 1. III. has
f x x f
f x
f x
f
a corner at 1. A I only
B II only
D
C III only
I and III only
Quick Quiz Sections 3.1 – 3.3
2. If the line normal to the graph of at the point 1, 2 passes through the point 1,1 , then which of the following gives the value of 1 ?
A 2 B 2
1 C
2 1
D 2
E 3
f
f
Quick Quiz Sections 3.1 – 3.3
2. If the line normal to the graph of at the point 1, 2 passes through the point 1,1 , then which of the following gives the value of 1 ?
A 2 B 2
1 C
1
2
E
D
3
2
f
f
Quick Quiz Sections 3.1 – 3.3
2 2 2 2 4 3 3. Find if .Quick Quiz Sections 3.1 – 3.3
2 2 2 2 1 4 3 3. Find if .3.4
Quick Review
2
In Exercises 1-10, answer the questions about the graph of the quadratic function 16 160 256 by analyzing the equation
algebraically. Then support your answers graphically. 1. Does the graph open u
y x x
pward or downward? 2. What is the -intercept?
3. What are the -intercepts?
4. What is the range of the function?
5. What point is the vertex of the parabola?
Quick Review
0 2
2
6. At what -values does 80? 7. For what -value does 100? 8. On what interval is 0?
3 3
9. Find lim . 10. Find at 7.
h
x f x
dy x
dx dy dx
f h f
h d y
x dx
Quick Review Solutions
2
In Exercises 1-10, answer the questions about the graph of the quadratic function 16 160 256 by analyzing the equation
algebraically. Then support your answers graphically. 1. Does the graph open u
y x x
Downward intercept 256
intercepts 2,8
pward or downward? 2. What is the -intercept?
3. What are the -intercepts?
4. What is the range of the function? ( ,144
5. What point is the vertex of the parabola? 5 ]
y x
y x
Quick Review Solutions
0 2 26. At what -values does 80? 7. For what -value does 100? 8. On what interval is 0?
3 3 9. Find 3,7 15 8 ,5 64 32 lim .
10. Find at 7.
h
x f x
dy x
dx dy dx
f h f
What you’ll learn about
Instantaneous Rates of change
Motion Along a Line
Sensitivity to Change
Derivatives in Economics
… and why
Instantaneous Rates of Change
0
The instantaneous rate of changeof with respect to at is the derivative
lim provided the limit exists.
When we say rate of change, we mean instantaneous rate of change. h
f x a
f a h f a f a
h
Example
Instantaneous Rates of Change
2
If the area of a square as a function of the radius is ,
a Find the rate of change of the area with respect to the radius . b Evaluate the rate of change of when 4.
A r
A r
A r
a The rate of change is the derivative 2 b The rate of change when 4 is 2 4 8
dA
r dr
r
Motion Along a Line
Suppose that an object is moving along a coordinate line so that we know its position s on that line as a function of time :
The of the object over the time interval from to is
t s f t
t t t
displacement
.
The of the object over that time interval is displacement
. travel time
av
s f t t f t
f t t f t s
v
t t
Instantaneous Velocity
0
The instantaneous velocity is the derivative of the position function with respect to time. At time the velocity is
lim .
t
s f t t
f t t f t ds
v t
dt t
Speed
Speed is the absolute value of velocity. Speed v t ds
dt
Acceleration
22Acceleration is the derivative of velocity with respect to time. If a body's velocity at time is then the body's
acceleration at time is .
ds t v t
dt dv d s t a t
dt dt
Free-fall Constants (Earth)
2 2
2
2 2
2
ft 1
English Units 32 , 32 16
sec 2
m 1
Metric Units 9.8 , 9.8 4.9
sec 2
g s t t
g s t t
Example
Finding Velocity
2
A projectile is shot upward from the surface of the earth and reaches a height of 4.9 120 meters after seconds.
Find the velocity after 5 seconds.
s t t t
9.8 120
when 5, 9.8 5 120 71 meters per second
ds
v t
dt
t v
Sensitivity to Change
Derivatives in Economics
Economists have a specialized vocabulary for rates of change and derivatives. They call them . In a manufacturing operation, the is a function of , the number of unitx
marginals
cost of production c(x) s
produced. The is the rate of change of cost with respect to the level of production, so it is .
Sometimes the marginal cost of production is loosely defined to be th
dc dx
marginal cost of production
Example
Derivatives in Economics
2Suppose that the dollar cost of producing washing machines is 2000 100 0.1 .
Find the marginal cost of producing 100 washing machines.
x c x x x
2
The marginal cost is 2000 100 0.1 100 0.2
The marginal cost of producing 100 machines is 100 0.2 100 $80
d
c x x x x
dx
3.5
Quick Review
1. Convert 135 degrees to radians.
2. Convert 1.7 radians to degrees.
3. Find the exact value of sin without a calculator. 3
4. State the domain and the range of the cosine function.
5. State the domain
Quick Review
2
3 2
6. If sin 1, what is cos ?
7. If tan 1, what are two possible values of sin ? 1 cos sin
8. Verify the identity: . 1 cos
9. Find an equation of the line tangent to the curve 2 7 10 at the point 3,
a a
a a
h h
h h h
y x x
3 2
1 . 10. A particle moves along a line with velocity
2 7 10 for time 0. Find the acceleration of the particle at 3.
v t t t
t
Quick Review Solutions
1. Convert 135 degrees to radians. 2. Convert 1.7 radians to degrees. 3. Find the exact va
3
2.356 4
97.403
3 2
lue of sin without a calculator. 3
4. State the domain and the range of the cosine function.
D
5. State the domain and the range of the tangent funct
omain: all reals Range: [-1,1]
Domain: odd integer Range:
ion.
all reals 2
k
Quick Review Solutions
2 2
2
6. If sin 1, what is cos ?
7. If tan 1, what are two possible values of sin ?
1 cos sin
8. Verify the identi
0
1 2
1 cos
Multiply by and use the identity
ty: .
1 cos
9. Find an equa
1 cos sin 1 os i c t a a a a h h
h h h
h h h h
3 2 3 2on of the line tangent to the curve 2 7 10 at the point 3,1 .
10. A particle moves along a line with velocity
2 7 10 for time 0. Find the acceleration
12 35
y x x
v t t t
What you’ll learn about
Derivative of the Sine Function
Derivative of the Cosine Function
Simple Harmonic Motion
Jerk
Derivatives of Other Basic Trigonometric Functions
… and why
Derivative of the Sine Function
The derivative of the sine is the cosine.
sin cos
d
x x
Derivative of the Cosine Function
The derivative of the cosine is the negative of the sine. cos sin
d
x x
Example
Finding the Derivative of the
Sine and Cosine Functions
sin
Find the derivative of . cos 2 x x
2 2 2 2 2 2 2 2 quotient rule 2 2sin cos 1
cos 2 sin sin cos 2 cos 2
cos 2 cos sin sin cos 2
cos 2cos sin cos 2
sin cos 2cos
cos 2 x x
d d
x x x x
dy dx dx
dx x
x x x x
x
x x x
x
x x x
x
Simple Harmonic Motion
Example
Simple Harmonic Motion
A weight hanging from a spring bobs up and down with position function 3sin in meters, in seconds . What are its velocity and acceleration at time ?
s t s t
t
3sin
3cos 3sin
s t
ds
v t
dt dv
a t
dt
Jerk
33is the derivative of acceleration. If a body's position at time is
.
t
da d s j t
dt dt
Derivative of the Other Basic
Trigonometric Functions
2 2
tan sec cot csc sec sec tan csc csc cot
d
x x
dx d
x x
dx d
x x x
dx d
x x x
dx
Example
Derivative of the Other Basic
Trigonometric Functions
Find the equation of a line tangent to y x cos at x x1.
cos
cos sin cos 1
Evaluate when 1
1 .8414709848 .5403023059 .3011686789 1 .8414709848 .5403023059 .3011686789 When 1, 1 cos1 .5403023059
The equation of the tangent line is .54030
y x x
d
m x x x x x
dx
m x
m m
x y
y
Example
Derivative of the Other Basic
Trigonometric Functions
cos
y x x
.3012 .841
3.6
Quick Review
2
In Exercises 1-5, let sin , 1, and 7 . Write a simplified expression for the composite function.
1. 2.
3. 4.
5.
f x x g x x h x x
f g x f g h x
g h x h g x
g x f
h x
Quick Review
2
2 2
4
In Exercises 6-10, let cos , 2, and
3 . Write the given function as a composite of two or more of , , and . For example, cos3 is .
6. cos 2 7. 3cos 2
8. 3 cos 6 9. cos 27
10. cos 2 3
f x x g x x h x x
f g h x f h x
x x
x x
Quick Review Solutions
2 2 2 2 2 2In Exercises 1-5, let sin , 1, and 7 . Write a simplified expression for the composite function.
sin
1. 2.
3. 4
1 sin 49 1
49 1 7 7
1 in 7 5. s . x x
f x x g x x h x x
f g x f g h x
g h x x h x
x
g x
g x f
h x x
Quick Review Solutions
2 2 2In Exercises 6-10, let cos , 2, and
3 . Write the given function as a composite of two or more of , , and . For example, cos3 is .
6. cos 2 7. 3cos 2
8. 3 cos 6 9. cos 27
f x x g x x h x x
f g h x f h x
x g f x g h f x
h g
x
x f x
4 210. cos 2 3
x
x
f h h x
What you’ll learn about
Derivative of a Composite Function
“Outside-Inside” Rule
Repeated Use of the Chain Rule
Slopes of Parametrized Curves
Power Chain Rule
… and why
The chain rule is the most widely used differentiation
Rule 8 The Chain Rule
If is differentiable at the point , and is differentiable at , then the composite function
is differentiable at , and
f u g x g
x
f g x f g x x
f g x f g x g x
Example
Derivatives of Composite
Functions
An object moves along the axis so that its position at any time 0 is given by sin 5 3 . Find the velocity of the object as a function of .
x
t s t t
t
sin 5 3 is a composite function: sin and 5 3 cos and 5
By the Chain Rule cos 5
5cos 5 3
s t t s u u t
ds du
u
du dt
“Outside-Inside” Rule
It sometimes helps to think about the Chain Rule this way:
If , then .
In words, differentiate the "outside" function and evaluate it at the "inside" function ( ) left alone; then
dy
y f g x f g x g x
dx
f g x
Example
“Outside-Inside” Rule
4
Differentiate cos 3x 2 with respect to .x
4 4 3
derivative inside inside left alone of inside
3 4
cos 3 2 sin 3 2 12
12 sin 3 2
d
x x x
dx
x x
Example
Repeated Use of the Chain Rule
3 2
Sometimes the chain rule needs to be used more than once to find a derivative.
Find the derivative of y sin .x
1
2 2
3 3
2
2 3 2
2
2 3 2
sin sin 1
sin cos 2
3 2
sin cos 3
y x x
dy
x x x
dx dy
x x x
dx
Slopes of Parametrized Curves
A parametrized curve , is if and are differentiable at .
x t y t differentiable at t
Finding dy/dx Parametrically
If all three derivatives exist and dx 0,
dt dy
dy dt dx dx
dt
Power Chain Rule
If is a differentiable function of , and is a differentiable function of , then substituting into the Chain Rule formula leads to the formula
f u u
x y f u
dy dy du dx du dx
d du
f u f u
dx dx
Quick Quiz
Sections 3.4 – 3.6
4 3
3 3
3
You should solve the following problems without using a graphing calculator. 1. Which of the following gives for sin 3 ?
A 4sin 3 cos 3 B 12sin 3 cos 3 C 12sin 3 cos 3 D 12sin 3
E 12sin 3 cos 3
dy
y x
dx
x x
x x
x x
x
x x
Quick Quiz
Sections 3.4 – 3.6
4 3
3 3
3
You should solve the following problems without using a graphing calculator. 1. Which of the following gives for sin 3 ?
A 4sin 3 cos 3
C 12sin 3 cos 3 D 12sin 3
E 12s
B 12sin 3 cos 3
in 3 cos 3 x
dy
y x
dx
x x
x x
x
x x
x
Quick Quiz
Sections 3.4 – 3.6
2 2 2 2 2
2. Which of the folllowing gives for cos tan ? A cos 2sec tan
B cos 2sec tan C sin sec
D cos sec tan E cos sec tan
y y x x
x x x
x x x
x x
x x x
x x x
Quick Quiz
Sections 3.4 – 3.6
2 2 2 2 2 2A cos 2sec ta
. Which of the folllowing gives for cos tan ?
B cos 2sec tan C sin sec
D cos sec tan E cos se
n
c tan
y y x x
x x x
x x
x x x
Quick Quiz
Sections 3.4 – 3.6
3. Which of the following gives for the parametric curve 3sin , 2cos ?
3
A cot
2 3 B cot
2 2 C tan
3 2 D tan
3
dy dx
x t y t
t
t
t
t
Quick Quiz
Sections 3.4 – 3.6
3. Which of the following gives for the parametric curve 3sin , 2cos ?
3
A cot
2 3 B cot
2
2 D tan
3 E t
2 C tan
3
an
dy dx
x t y t
t
t
t
t t
3.7
Quick Review
1 2
2
2 2
2 2
2 2
In Exercises 1-5, sketch the curve defined by the equation and find two functions and whose graphs will combine to give the curve.
1. 0
2. 4 9 36 3. 4 0 4. 9 5.
y y
x y
x y
x y
x y
2 2 2 3
Quick Review
2
2 2
In Exercises 6-8, solve for in terms of and . 6. 2 4
7. sin cos
8.
y y x
x y xy x y
y x x x xy y
x y y y x y
Quick Review
3
3 2 3
In Exercises 9 and 10, find an expression for the function using rational powers rather than radicals.
9.
10.
x x x
x x x
Quick Review Solutions
1 2 2 2 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 , 2 2 9 ,
1. 0
2. 4 9 36 3. 4 0 4. 9 5
9
3 3
,
2 2
9 , 9
2 3 , 2 3
. 2 3
y x y x
y x y x
x x y x y x y x y x y x y y
y x y x
y x x y x x
x 1 2
In Exercises 1-5, sketch the curve defined by the equation and find two functions and whose graphs will combine to give the curve.
Quick Review Solutions
In Exercises 6-8, solve for in terms of and .y y x
2
2
2 2 2
2
6. 2 4
7. sin cos
8.
4 2
cos sin
x y xy x y
y x x x x
x y xy
y
x
y x x
y
x x
xy y
y y
x y y x y
Quick Review Solutions
In Exercises 9 and 10, find an expression for the function using rational powers rather than radicals.
3 56 2
5 1
3
2 3
2
3
6
9.
10.
x
x x x
x x
x
x x
x
What you’ll learn about
Implicitly Defined Functions
Lenses, Tangents, and Normal Lines
Derivatives of Higher Order
Rational Powers of Differentiable Functions
… and why
Implicitly Defined Functions
An important problem in Calculus is how to find the slope when the function can't conveniently be solved for . Instead, is treated as a differentiable function of and both sides of the equationy y
x are
differentiated with respect to , using the appropriate rules for sums, products, quotients and the Chain rule. Then solve for in terms of and together to obtain a formula that calculates
x
dy
x dx
y
the slope at any point , on the graph.
The process by which we find is called . The phrase derives from the fact that the equation
x y
dy
Example
Implicitly Defined Functions
2
Find dy if 3y 2y 5x
dx
2
2 2
To find differentiate both sides of the equation with respect to , treating as a differentiable function of and applying the Chain Rule.
3 2 5
3 3
6 2 5
2 2
dy dx
x y x
y y x
d d dy
y y
dx dy dx
dy dy y
d d dy
Implicit Differentiation Process
1. Differentiate both sides of the equation with respect to . 2. Collect the terms with on one side of the equation. 3. Factor out .
4. Solve for .
x dy
dx dy
Lenses, Tangents and Normal Lines
In the law that describes how light changes direction as it enters a
lens, the important angles are the angles the light makes with the
line perpendicular to the surface of the lens at the point of entry
(angles
A
and
B
in Figure 3.50). This line is called the
normal to
the surface
at the point of entry. In a profile view of a lens, the
normal is a line perpendicular to the tangent to the profile curve
at the point of entry.
Example
Lenses, Tangents and Normal
Lines
4 2 2 2
Find the equations of the tangent and normal lines to the graph 2 2
given by 0 at the point , . 2 2
x x y y
2 24 2 2 2
3 2 2
treat as a product
2 2 2
2 2
Differentiate implicitly as
0
4 2 2 2 0
2 1 2 2
2 2
x y
d d d
x x y y
dx dx dx
dy dy
x x y xy y
dx dx
dy
y x x x y
dx
