Transcript Chapter 3 Derivatives - Mr. Wojcik's Classes / FrontPage

Chapter 3 Derivatives
3.1 Derivative of a Function
The derivative of the function f with respect to the variable x is the
function f   x  whose value at x is
f   x   lim
h 0
f  x  h  f  x
h
provided the limit exists.
The domain of f , the set of points in the domain of f for which the limit
exists, may be smaller than the domain of f . If f   x  exists, we say that f
has a derivative (is differentiable) at x. A function that is differentiable
at every point in its domain is a differentiable function.
Differentiate f  x   x 2
There are many ways to denote the derivative of a function y  f ( x).
Besides f '( x), the most common notations are:
y
" y prime"
Nice and brief, but does not name
the independent variable.
dy
dx
" dy dx " or "the derivative
of y with respect to x "
Names both variables and
uses d for derivative.
df
dx
" df dx " or "the derivative
of f with respect to x "
Emphasizes the function's name.
d
f  x  " d dx of f at x " or "the
dx
derivative of f at x "
Emphasis that differentiation
is an operation performed on f .
The derivative of the function f at the point where x  a is the limit
f   a   lim
x a
f  x  f a
xa
provided the limit exists.
Differentiate f ( x) 
x using the alternate definition.
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.
We estimate the slope of the graph of f in y-units per xunit at frequent intervals. We then plot the estimates in a
coordinate plane with the horizontal axis in x-units and the
vertical axis in slope units.
Look at page 103
Show that the following function has left-hand and right-hand
derivatives at x  0, but no derivative there.
  x,
y
 x,
Left-hand derivative:
lim
 0  h  0
h 0 
 lim
h 0
h
h
 1
h
x0
x0
Right-hand derivative:
0  h  0

lim
h 0 
h
h
 lim 1
h 0 h
The derivatives are not equal at x  0. The function does not
have a derivative at 0.
p.105 (1-19)odd, (26-28)
3.2 Differentiability
How f’(a) Might Fail to Exist
A function will not have a derivative at a point P  a, f  a  
where the slopes of the secant lines,
f  x  f a
xa
fail to approach a limit as x approaches a.
The next figures illustrate four different instances where this occurs.
For example, a function whose graph is otherwise smooth will fail to
have a derivative at a point where the graph has:
1. a corner, where the one-sided derivatives differ;
f  x  x
2. 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
2
3
3. A vertical tangent, where the slopes of the secant lines approach
either  or - from both sides;
f  x  3 x
4.
a discontinuity (which will cause one or both of the one-sided
derivatives to be nonexistent).
1, x  0
U  x  
1, x  0
Show that the function is not differentiable at x  0.
 x3 , x  0
f  x  
 4 x, x  0
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
differentiability.
What is linearization?
Let’s discuss calculator derivatives.
Differentiability Implies Continuity
If f has a derivative at x  a, then f is continuous at x  a.
The converse is not a true statement.
Intermediate Value Theorem for
Derivatives
If a and b are any two points in an interval on which f is
differentiable, then f  takes on every value between
f   a  and f   b  .
p. 114 (1-37) odd
3.3 Rules for Differentiation
If f is the function with the constant value c, then
df d
 c  0
dx dx
This means that the derivative of every constant function
is the zero function.
If n is a positive integer, then
d n
x   nx n 1

dx
The Power Rule says:
To differentiate x n , multiply by n and subtract 1 from the exponent.
If u is a differentiable function of x and c is a constant, then
d
du
cu

c
 
dx
dx
This says that if a differentiable function is multiplied by a constant,
then its derivative is multiplied by the same constant.
If u and v are differentiable functions of x, then their sum and differences
are differentiable at every point where u and v are differentiable. At such points,
d
du dv
u  v   
dx
dx dx.
3
Differentiate the polynomial y  x  2 x  x  19
4
dy
That is, find
.
dx
4
2
Does the curve y  x 4  8 x 2  2 have any horizontal tangents?
If so, where do they occur?
Verify you result by graphing the function.
The product of two differentiable functions u and v is differentiable, and
d
dv
du
uv

u

v
 
dx
dx
dx
The derivative of a product is actually the sum of two products.
Find f   x  if f  x    x3  4  x 2  3
At a point where v  0, the quotient y 
u
of two differentiable
v
functions is differentiable, and
du
dv
u
d u
dx
dx

 
dx  v 
v2
Since order is important in subtraction, be sure to set up the
v
numerator of the Quotient rule correctly.
Find f   x 
x3  4
if f  x   2
x 3
If n is a negative integer and x  0, then
d n
x   nx n 1 .

dx
This is basically the same as Rule 2 except now n is negative.
Find an equation for the line tangent to the curve y 
1
at the point 1,1 .
x
dy
is called the first derivative of y with respect to x.
dx
The first derivative may itself be a differentiable function of x. If so,
The derivative y  
dy  d  dy  d 2 y
its derivative,
y  
   2 ,
dx dx  dx  dx
is called the second derivative of y with respect to x. If y 
 y double prime  is differentiable, its derivative,
dy  d 3 y
y  
 3,
dx dx
is called the third derivative of y with respect to x.
Find an equation for the line tangent to the curve
x2  3
y
2x
at the point (1,2).
Find the first four derivatives of
y  x3  5 x 2  2
p.124 (1-47) odd
3.4 Velocity and Rates of Change
The  instantaneous  rate of changeof f with respect to x at a
is the derivative
f   a   lim
h 0
f a  h  f a
h
provided the limit exists.
When we say rate of change, we mean instantaneous rate of change.
If the area of a circle as a function of the radius is A   r 2 ,
a 
b
Find the rate of change of the area A with respect to the radius r.
Evaluate the rate of change of A when r  4.
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 t:
s  f t 
The displacement of the object over the time interval from
t to t  t is s  f  t  t   f  t  .
The average velocity of the object over that time interval is
displacement s f  t  t   f  t 
vav 
 
.
travel time
t
t
The  instantaneous  velocity is the derivative of the position
function s  f  t  with respect to time. At time t the velocity is
v t  
f  t  t   f  t 
ds
 lim
.

t

0
dt
t
Speed is the absolute value of velocity.
Speed  v  t  
ds
dt
A projectile is shot upward from the surface of the earth and reaches
a height of s   4.9t 2  120t meters after t seconds.
Find the velocity after 5 seconds.
Look at page 129 on how to read a velocity graph.
Acceleration is the derivative of velocity with respect to time.
ds
then the body's
dt
dv d 2 s
acceleration at time t is a  t    2 .
dt dt
If a body's velocity at time t is v  t  
English Units
Metric Units
ft
g  32 2 ,
sec
m
g  9.8 2 ,
sec
1
s   32  t 2 16t 2
2
1
s   9.8  t 2  4.9t 2
2
When a small change in x produces a large change in the value of a function
f(x), we say that the function is relatively sensitive to changes in x. The
derivative f’(x) is a measure of this sensitivity.
Economists have a specialized vocabulary for rates of change and
derivatives. They call them marginals. In a manufacturing operation,
the cost of production c(x) is a function of x, the number of units
produced. The marginal cost of production is the rate of change of
dc
cost with respect to the level of production, so it is
.
dx
Sometimes the marginal cost of production is loosely defined to be
the extra cost of producing one more unit.
Suppose that the dollar cost of producing x washing machines
is c  x   2000  100 x  0.1x 2 .
Find the marginal cost of producing 100 washing machines.
p. 135 (1 -39) odd
3.5 Derivatives of Trigonometric
Functions
The derivative of the sine is the cosine.
d
sin x  cos x
dx
Let’s prove this using the definition of the derivative.
The derivative of the cosine is the negative of the sine.
d
cos x   sin x
dx
Find the derivative of
sin x
.
 cos x  2 
Find the derivative of y  x sin x
2
The motion of a weight bobbing up and down on the end of
a string is an example of simple harmonic motion.
A weight hanging from a spring bobs up and down with position
function s  3sin t  s in meters, t in seconds  . What are its velocity
and acceleration at time t ?
d
tan x  sec 2 x
dx
d
cot x   csc 2 x
dx
d
sec x  sec x tan x
dx
d
csc x   csc x cot x
dx
Let’s derive the formula for tangent.
Find the equation of a line tangent to y  x cos x at x 1.
Find yn if
y  sec x
p. 146 (1-41) odd
3.6 Chain Rule
If f is differentiable at the point u  g  x  , and g is
differentiable at x, then the composite function
f
g  x   f  g  x   is differentiable at x, and
f
g   x   f   g  x    g   x 
An object moves along the x  axis so that its position at any
time t  0 is given by s  t   sin  5t  3 . Find the velocity of the
object as a function of t.
Differentiate cos  3x 4  2  with respect to x.
Sometimes the chain rule needs to be used
more than once to find a derivative.
Find the derivative of y  3 sin x 2 .
Power Chain Rule
If f is a differentiable function of u, and u is a differentiable
function of x, then substituting y  f  u  into the Chain Rule
formula
dy dy du
 
dx du dx
leads to the formula
d
du
f u   f  u 
dx
dx
Ex: (a) Find the slope of the line tangent to the curve
where x 
y  sin5 x at the point

3
(b) Show that the slope of every line tangent to the curve y 
1
1  2x 
3
is positive.
p. 153 (1-39, 53-69)odd
skip (57), also do (56, 58)
3.7 Implicit Differentiation
Find
dy
if 3 y 2  2 y  5 x
dx
Implicit Differentiation Process
1. Differentiate both sides of the equation with respect to x.
dy
2. Collect the terms with
on one side of the equation.
dx
dy
3. Factor out
.
dx
dy
4. Solve for
.
dx
Find the equations of the tangent and normal lines to the graph
 2
2
given by x  x y  y  0 at the point 
,
.
2 
 2
4
2
2
2
If n is any rational number, then
d n
x  nx n1.
dx
If n  1, then the derivative does not exist at x  0.
d
dx
 x
d 4/5
x 

dx
d
1/ 5
cos
x


dx
p.162 (1-57) odd and 54
3.8 Derivatives of Inverse
Trigonometric Functions
Derivatives of Inverse Functions
dy
dx
is differentiable
If f is differentiable at every point of an interval I and
is never zero on I , then f has an inverse and f 1
at every point on the interval f  I  .
Let’s go the exploration of page 166
If u is a differentiable function of x with u  1, we apply the
Chain Rule to get
d
1 du
sin 1 u 
, u  1.
2
dx
1  u dx
How did we get this as a result?
Identity Functions
cos 1 x 
cot 1 x 
csc 1 x 

2

2

2
 sin 1 x
1
sec 1 x  cos 1  
x
 tan 1 x
cot 1 x 

2
 tan 1 x
1
csc 1 x  sin 1  
x
 sec 1 x
Rules for differentiation
d
1
sin 1 x  

dx
1  x2
d
1
csc 1 x   

dx
x x2 1
d
1
cos 1 x   

dx
1  x2
d
1
1
sec
x



dx
x x2 1
d
1
1
tan
x


 1  x2
dx
d
1
cot 1 x   

dx
1  x2
Ex: d  sin 1 x 2 
dx
Ex: A particle moves along the x-axis so that its position at any time t  0
Is x(t )  tan 1 t. What is the velocity of the particle when t = 16?
d
1
4
sec
5
x


Ex: dx
p. 170 (1-33) odd
3.9 Derivative of Exponential and
Logarithmic Functions
d x
e 

dx
If u is a differentiable function of x, then
d u u du
e e
dx
dx
dy
x3
Find
if y  e
dx
If u is a differentiable function of x and for a  0 and a  1,
d u
du
u
a

a
ln
a
 
dx
dx
Ex: At what point on the graph of the function y = 3t – 3 does the
tangent line have a slope of 21?
If u is a differentiable function of x and u  0,
d
1 du
ln u 
dx
u dx
How can we show this?
Find
dy
if y  ln  3 x 2 
dx
If u is a differentiable function of x and u  0,
d
1 du
log a u 
dx
u ln a dx
If u is a positive differentiable function of x and n is a
differentiable function of x, and
d n
du
u  nu n1 .
dx
dx
d
log a a sin x 

dx
If y   5 x  , find
3
4
dy
dx.
Sometimes the properties of logarithms can be used to
simplify the differentiation process, even if logarithms
themselves must be introduced as a step in the process.
The process of introducing logarithms before differentiating
is called logarithmic differentiation.
Find
dy
for y  x cos x
dx
p.178 (1-55)odd