Differential calculus

DERIVATIVE
• Recall your concept of a slope;

What happens to the graph as the
change in x approaches 0?

Example
•Find an equation of the tangent line to
the parabola y = x2
-1 at point (2,3).
Sketch the parabola and show the
segment of the tangent line at (2,3).
Find the general formula for the slope
of the tangent line of this function.
What is the slope when x = 5?

2
3y x= −
( ) ( )2 2
0
3 3
lim
h
x h x
y
h→
+ − − −
′ =
2y x′ =
0
lim2
h
y x h
→
′ = +
3.1 Derivative of a Function

A function is differentiable if it has a
derivative everywhere in its domain.
It must be continuous and smooth.
Functions on closed intervals must
have one-sided derivatives defined
at the end points.
Derivative of a Function

To be differentiable, a function must be continuous
and smooth.
Derivatives will fail to exist at:
corner
( )f x x=
cusp
( )
2
3
f x x=
vertical tangent
( ) 3
f x x=
discontinuity
( )
1, 0
1, 0
x
f x
x
− <
=
≥
3.2 Differentiability

Most of the functions we study in calculus will be differentiable.
Differentiability

If f has a derivative at x = a, then f is continuous at x = a.
Since a function must be continuous to have a derivative,
if it has a derivative then it is continuous.
3.2 Differentiability

( )
1
2
f a′ =
( ) 3f b′ =
Intermediate Value Theorem for Derivatives
Between a and b, must take
on every value between and .
f ′
1
2
3
If a and b are any two points in an interval on which f is
differentiable, then takes on every value between
and .
f ′ ( )f a′
( )f b′
Differentiability

If the derivative of a function is its slope, then for a
constant function, the derivative must be zero.
( ) 0
d
c
dx
=
example: 3y =
0y′ =
The derivative of a constant is zero.
3.3 Rules for Differentiation

We saw that if , .
2
y x= 2y x′ =
This is part of a pattern.
( ) 1n nd
x nx
dx
−
=
examples:
( ) 4
f x x=
( ) 3
4f x x′ =
8
y x=
7
8y x′ =
power rule
Rules for Differentiation

( ) 1n nd
x nx
dx
−
=
Proof:
h
xhx
x
dx
d nn
h
n −+
=
→
)(
lim
0
h
xhhnxx
x
dx
d nnnn
h
n −+++
=
−
→
...
lim
1
0
h
hhnx
x
dx
d nn
h
n ++
=
−
→
...
lim
1
0
1
0
lim −
→
= n
h
n
nxx
dx
d

( )
d du
cu c
dx dx
=
examples:
1n nd
cx cnx
dx
−
=
constant multiple rule:
5 4 4
7 7 5 35
d
x x x
dx
= ⋅ =

(Each term is treated separately)
( )
d du
cu c
dx dx
=constant multiple rule:
sum and difference rules:
( )
d du dv
u v
dx dx dx
+ = + ( )
d du dv
u v
dx dx dx
− = −
4
12y x x= +
3
4 12y x′ = +
4 2
2 2y x x= − +
3
4 4
dy
x x
dx
= −
3.3 Rules for Differentiation

Find the horizontal tangents of:
4 2
2 2y x x= − + 3
4 4
dy
x x
dx
= −
Horizontal tangents occur when slope = zero.
3
4 4 0x x− =
3
0x x− =
( )2
1 0x x − =
( ) ( )1 1 0x x x+ − =
0, 1, 1x = −
Substituting the x values into the
original equation, we get:
2, 1, 1y y y= = =
(The function is even, so we
only get two horizontal
tangents.)

4 2
2 2y x x= − +
2y =
1y =

4 2
2 2y x x= − +
First derivative
(slope) is zero at:
0, 1, 1x = −
3
4 4
dy
x x
dx
= −

product rule:
( )
d dv du
uv u v
dx dx dx
= + Notice that this is not just the
product of two derivatives.
This is sometimes memorized as: ( )d uv u dv v du= +
( )( )2 3
3 2 5
d
x x x
dx
 + + 
( )5 3 3
2 5 6 15
d
x x x x
dx
+ + +
( )5 3
2 11 15
d
x x x
dx
+ +
4 2
10 33 15x x+ +
=( )2
3x + ( )2
6 5x + ( )3
2 5x x+ + ( )2x
4 2 2 4 2
6 5 18 15 4 10x x x x x+ + + + +
4 2
10 33 15x x+ +

product rule:
( )
d dv du
uv u v
dx dx dx
= +
Proof
h
xvxuhxvhxu
uv
dx
d
h
)()()()(
lim)(
0
−++
=
→
add and subtract u(x+h)v(x)
in the denominator
h
xvhxuxvhxuxvxuhxvhxu
uv
dx
d
h
)()()()()()()()(
lim)(
0
+−++−++
=
→
( ) ( )





 −++−++
=
→ h
xuhxuxvxvhxvhxu
uv
dx
d
h
)()()()()()(
lim)(
0
dx
du
v
dx
dv
uuv
dx
d
+=)(

quotient rule:
2
du dv
v u
d u dx dx
dx v v
−
 
= 
 
or 2
u v du u dv
d
v v
− 
= 
 
3
2
2 5
3
d x x
dx x
+
+
( )( ) ( )( )
( )
2 2 3
2
2
3 6 5 2 5 2
3
x x x x x
x
+ + − +
=
+

Higher Order Derivatives:
dy
y
dx
′ = is the first derivative of y with respect to x.
2
2
dy d dy d y
y
dx dx dx dx
′
′′ = = =
is the second derivative.
(y double prime)
dy
y
dx
′′
′′′ = is the third derivative.
( )4 d
y y
dx
′′′= is the fourth derivative.
We will learn
later what these
higher order
derivatives are
used for.

Suppose u and v are functions that are differentiable at
x = 3, and that u(3) = 5, u’(3) = -7, v(3) = 1, and v’(3)= 4.
Find the following at x = 3 :
)(.1 uv
dx
d
'')( vuuvuv
dx
d
+= 8)7)(1()3(5 =−+






v
u
dx
d
.2 2
''
v
uvvu
v
u
dx
d −
=





2
1
)4)(5()7)(1( −−
27−=






u
v
dx
d
.3 2
''
u
vuuv
u
v
dx
d −
=





2
5
)7)(1()4)(5( −−
25
27
=







hi
ho
dx
d
))((
)()()()(
hoho
hidhohodhi −

Consider a graph of displacement (distance traveled) vs. time.
time (hours)
distance
(miles)
Average velocity can be found by
taking:
change in position
change in time
s
t
∆
=
∆
t∆
s∆
A
B
( ) ( )
ave
f t t f ts
V
t t
+∆ −∆
= =
∆ ∆
The speedometer in your car does not measure average
velocity, but instantaneous velocity.
( )
( ) ( )
0
lim
t
f t t f tds
V t
dt t∆ →
+∆ −
= =
∆
(The velocity at one
moment in time.)
Velocity and other Rates of Change

Velocity is the first derivative of position.
Acceleration is the second derivative
of position.

Example: Free Fall Equation
21
2
s g t=
Gravitational
Constants:
2
ft
32
sec
g =
2
m
9.8
sec
g =
2
cm
980
sec
g =
21
32
2
s t= ⋅
2
16s t= 32
ds
V t
dt
= =
Speed is the absolute value of velocity.

Acceleration is the derivative of velocity.
dv
a
dt
=
2
2
d s
dt
= example:
32v t=
32a =
If distance is in: feet
Velocity would be in:
feet
sec
Acceleration would be in:
ft
sec
sec
2
ft
sec
=

Rates of Change:
Average rate of change =
( ) ( )f x h f x
h
+ −
Instantaneous rate of change = ( )
( ) ( )
0
lim
h
f x h f x
f x
h→
+ −
′ =
These definitions are true for any function.
( x does not have to represent time. )

For a circle: 2
A rπ=
2dA d
r
dr dr
π=
2
dA
r
dr
π=
Instantaneous rate of change of the area with
respect to the radius.
For tree ring growth, if the change in area is constant then dr
must get smaller as r gets larger.
2dA r drπ=

from Economics:
Marginal cost is the first derivative of the cost function, and
represents an approximation of the cost of producing one
more unit.

Example 13:
Suppose it costs: ( ) 3 2
6 15c x x x x= − +
to produce x stoves. ( ) 2
3 12 15c x x x′ = − +
If you are currently producing 10 stoves,
the 11th
stove will cost approximately:
( ) 2
10 3 10 12 10 15c′ = × − × +
300 120 15= − +
$195=
marginal cost
The actual cost is: ( ) ( )11 10C C−
( ) ( )3 2 3 2
11 6 11 15 11 10 6 10 15 10= − × + × − − × + ×
770 550= − $220= actual cost

Derivatives of Trigonometric Functions

π
2
π
0
2
π
−
π−
Consider the function ( )siny θ=
We could make a graph of the slope:
θ slope
1−
0
1
0
1−
Now we connect the dots!
The resulting curve is a cosine curve.
( )sin cos
d
x x
dx
=

h
xhx
x
dx
d
h
sin)sin(
limsin
0
−+
=
→
h
xxhhx
x
dx
d
h
sincossincossin
limsin
0
−+
=
→
h
xh
h
hx
x
dx
d
hh
cossin
lim
)1(cossin
limsin
00 →→
+
−
=
h
xhhx
x
dx
d
h
cossin)1(cossin
limsin
0
+−
=
→
Proof

h
xh
h
hx
x
dx
d
hh
cossin
lim
)1(cossin
limsin
00 →→
+
−
=
= 0 = 1
( )sin cos
d
x x
dx
=

h
xhx
x
dx
d
h
cos)cos(
limcos
0
−+
=
→
h
xxhhx
x
dx
d
h
cossinsincoscos
limcos
0
−−
=
→
h
xh
h
hx
x
dx
d
hh
sinsin
lim
)1(coscos
limcos
00 →→
−
−
=
h
xhhx
x
dx
d
h
sinsin)1(coscos
limcos
0
−−
=
→
Find the derivative of cos x

= 0 = 1
h
xh
h
hx
x
dx
d
hh
sinsin
lim
)1(coscos
limcos
00 →→
−
−
=
( )cos sin
d
x x
dx
= −

We can find the derivative of tangent x by using the
quotient rule.
tan
d
x
dx
sin
cos
d x
dx x
( )
2
cos cos sin sin
cos
x x x x
x
× − ×−
2 2
2
cos sin
cos
x x
x
+
2
1
cos x
2
sec x
( ) 2
tan sec
d
x x
dx
=

Derivatives of the remaining trig functions can
be determined the same way.
sin cos
d
x x
dx
=
cos sin
d
x x
dx
= −
2
tan sec
d
x x
dx
=
2
cot csc
d
x x
dx
= −
sec sec tan
d
x x x
dx
= ×
csc csc cot
d
x x x
dx
= − ×

Jerk A sudden change in acceleration
Definition Jerk
Jerk is the derivative of acceleration. If a body’s position
at time t is s(t), the body’s jerk at time t is
3
3
2
2
)(
dt
sd
dt
vd
dt
da
tj ===

Consider a simple composite function:
6 10y x= −
( )2 3 5y x= −
If 3 5u x= −
then 2y u=
6 10y x= − 2y u= 3 5u x= −
6
dy
dx
= 2
dy
du
= 3
du
dx
=
dy dy du
dx du dx
= ⋅
6 2 3= ⋅
Chain Rule

dy dy du
dx du dx
= ⋅Chain Rule:
example: ( ) sinf x x= ( ) 2
4g x x= − Find: ( ) at 2f g x′ =o
( ) cosf x x′ = ( ) 2g x x′ = ( )2 4 4 0g = − =
( ) ( )0 2f g′ ′⋅ ( ) ( )cos 0 2 2⋅ ⋅ 1 4⋅ 4=
Chain Rule
If is the composite of and ,
then:
f go ( )y f u= ( )u g x=
( ) ( ) atat xu g x
f g f g=
′ ′ ′= ⋅o )('))((' xgxgf •=

( )( ) ( )2
sin 4f g x x= −
( )2
sin 4y x= −
siny u=
2
4u x= −
cos
dy
u
du
= 2
du
x
dx
=
dy dy du
dx du dx
= ⋅
cos 2
dy
u x
dx
= ⋅
( )2
cos 4 2
dy
x x
dx
= − ⋅
( )2
cos 2 4 2 2
dy
dx
= − ⋅ ⋅
( )cos 0 4
dy
dx
= ⋅
4
dy
dx
=
Chain Rule

Here is a faster way to find the derivative:
( )2
sin 4y x= −
( ) ( )2 2
cos 4 4
d
y x x
dx
= − ⋅ −
( )2
cos 4 2y x x= − ⋅
Differentiate the outside function...
…then the inside function
At 2, 4x y= =
Chain Rule

( )2
cos 3
d
x
dx
( )
2
cos 3
d
x
dx
  
( ) ( )2 cos 3 cos 3
d
x x
dx
⋅  
( ) ( ) ( )2cos 3 sin 3 3
d
x x x
dx
⋅− ⋅
( ) ( )2cos 3 sin 3 3x x− ⋅ ⋅
( ) ( )6cos 3 sin 3x x−
The chain rule can be used
more than once.
(That’s what makes the
“chain” in the “chain rule”!)
Chain Rule

Derivative formulas include the chain rule!
1n nd du
u nu
dx dx
−
= sin cos
d du
u u
dx dx
=
cos sin
d du
u u
dx dx
= − 2
tan sec
d du
u u
dx dx
=
etcetera…
Chain Rule

Chain Rule
Find
)3cos( 2
xxy += )16)(3sin( 2
++−= xxx
dx
dy
))sin(cos(xy =
)24(cos 33
xxy +=
)sin)(cos(cos xx
dx
dy
−=
)212))(24sin()(24(cos3 2332
++−+= xxxxx
dx
dy
))24sin()(24(cos)636( 3322
xxxxx
dx
dy
+−++=
dx
dy

The chain rule enables us to find the slope of
parametrically defined curves:
dy dy dx
dt dx dt
= ⋅
dy
dydt
dx dx
dt
=
The slope of a parametrized
curve is given by:
dy
dy dt
dxdx
dt
=
Chain Rule

2 2
1x y+ =
This is not a function,
but it would still be
nice to be able to find
the slope.
2 2
1
d d d
x y
dx dx dx
+ = Do the same thing to both sides.
2 2 0
dy
x y
dx
+ =
Note use of chain rule.
2 2
dy
y x
dx
= −
2
2
dy x
dx y
−
= dy x
dx y
= −
Implicit Differentiation

2
2 siny x y= +
2
2 sin
d d d
y x y
dx dx dx
= +
This can’t be solved for y.
2 2 cos
dy dy
x y
dx dx
= +
2 cos 2
dy dy
y x
dx dx
− =
( ) 22 cos
dy
xy
dx
=−
2
2 cos
dy x
dx y
=
−
This technique is called
implicit differentiation.
1 Differentiate both sides w.r.t. x.
2 Solve for .
dy
dx

Implicit Differentiation Process
1. Differentiate both sides of the equation with respect to x.
2. Collect the terms with dy/dx on one side of the equation.
3. Factor out dy/dx .
4. Solve for dy/dx .

Find the equations of the lines tangent and normal to the
curve at .
2 2
7x xy y− + = ( 1,2)−
2 2
7x xy y− + =
2 2 0
dydy
x yx y
dxdx
 − + =+  
Note product rule.
2 2 0
dy dy
x x y y
dx dx
− − + =
( ) 22
dy
y xy x
dx
= −−
2
2
dy y x
dx y x
−
=
−
( )
( )
2 2 1
2 2 1
m
− −
=
⋅ − −
2 2
4 1
+
=
+
4
5
=

Find the equations of the lines tangent and normal to the
curve at .
2 2
7x xy y− + = ( 1,2)−
4
5
m =
tangent:
( )
4
2 1
5
y x− = +
4 4
2
5 5
y x− = +
4 14
5 5
y x= +
normal:
( )
5
2 1
4
y x− = − +
5 5
2
4 4
y x− = − −
5 3
4 4
y x= − +

Find if .
2
2
d y
dx
3 2
2 3 7x y− =
3 2
2 3 7x y− =
2
6 6 0x y y′− =
2
6 6y y x′− = −
2
6
6
x
y
y
−
′ =
−
2
x
y
y
′ =
2
2
2y x x y
y
y
′⋅ −
′′ =
2
2
2x x
y y
y y
′′ ′= −
2 2
2
2x x
y
y
x
yy
′′ = − ⋅
4
3
2x x
y
y y
′′ = −
Substitute
back into the
equation.
y′

Rational Powers of Differentiable Functions
Power Rule for Rational Powers of x
If n is any rational number, then
1−
= nn
nxx
dx
d

Proof: Let p and q be integers with q > 0.
q
p
xy =
pq
xy =
Raise both sides to the q power
Differentiate with respect to x
11 −−
= pq
px
dx
dy
qy Solve for dy/dx

1
1
−
−
= q
p
qy
px
dx
dy Substitute for y
1/
1
)( −
−
= qqp
p
xq
px
dx
dy
Remove parenthesis
qpp
p
qx
px
dx
dy
/
1
−
−
= Subtract exponents
q
px
dx
dy qppp )/(1 −−−
=
1)/( −
= qp
x
q
p
dx
dy

Answer
4�3
−3�2
�
�3+�3

Look at the graph of
x
y e=
The slope at x = 0
appears to be 1.
If we assume this to be
true, then:
0 0
0
lim 1
h
h
e e
h
+
→
−
=
definition of derivative
Derivatives of Exponential and Logarithmic
Functions

Now we attempt to find a general formula for the
derivative of using the definition.
x
y e=
( ) 0
lim
x h x
x
h
d e e
e
dx h
+
→
−
=
0
lim
x h x
h
e e e
h→
⋅ −
=
0
1
lim
h
x
h
e
e
h→
 −
= ⋅ 
 
0
1
lim
h
x
h
e
e
h→
 −
= ⋅  
 
1x
e= ⋅
x
e=
This is the slope at x = 0, which
we have assumed to be 1.
Functions

75
– More Derivatives
4) Exponents
If y = bx
y’ = bx
ln(b)
Therefore
If y = ex
y’ = ex

77
– More Exciting Derivatives
2) Natural Logs
If y=ln(x),
y’ = 1/x
-chain rule may apply
If y=ln(x2
)
y’ = (1/x2
)2x = 2/x

x
e is its own derivative!
If we incorporate the chain rule: u ud du
e e
dx dx
=
We can now use this formula to find the derivative of
x
a
Derivatives of Exponential and
Logarithmic Functions

( )xd
a
dx
( )ln x
ad
e
dx
( )lnx ad
e
dx
( )ln
lnx a d
e x a
dx
⋅
Incorporating the chain rule:
( ) lnu ud du
a a a
dx dx
=
aaa
dx
d xx
ln=

So far today we have:
u ud du
e e
dx dx
= ( ) lnu ud du
a a a
dx dx
=
Now it is relatively easy to find the derivative of .ln x
3.9 Derivatives of Exponential and

lny x=
y
e x=
( ) ( )yd d
e x
dx dx
=
1y dy
e
dx
=
1
y
dy
dx e
=
1
ln
d
x
dx x
=
1
ln
d du
u
dx u dx
=

To find the derivative of a common log function, you
could just use the change of base rule for logs:
log
d
x
dx
ln
ln10
d x
dx
=
1
ln
ln10
d
x
dx
=
1 1
ln10 x
= ⋅
The formula for the derivative of a log of any base
other than e is:
1
log
ln
a
d du
u
dx u a dx
=

u ud du
e e
dx dx
= ( ) lnu ud du
a a a
dx dx
=
1
log
ln
a
d du
u
dx u a dx
=
1
ln
d du
u
dx u dx
=

84
– More Derivatives
4b) Exponents and chain rule
If y = bkx
y’ = bkx
ln(b)k
Or more generally:
If y = bg(x)
y’ = bg(x)
ln(b) X dg(x)/dx

85
2.1.2 – More Derivatives
4b) Exponents and chain rule
If y = 52x
y’ = 52x
ln(5)2
Or more complicated:
If y = 5sin(x)
y’ = 5sin(x)
ln(5) * cos(x)

Functions
x
ey 2
=
2
3x
y =
3
ln xy =
Find y’
x
ey 2
2'=
)2)(3ln(3'
2
xy x
=
x
x
x
y
3
)3(
1
' 2
3
==

Functions
Logarithmic differentiation
Used when the variable is in the base and the exponent
y = xx
ln y = ln xx
ln y = x ln x
x
x
x
dx
dy
y
ln
11
+





=
( )xy
dx
dy
ln1+=
( )xx
dx
dy x
ln1+=

Differentiate
Differentiate
Differentiate

Extrema of a Function
In calculus, much effort is devoted to determining the behavior of a
function f on an interval I.
Does f have a maximum value on I? Does it have a minimum value?
Where is the function increasing? Where is it decreasing?
In this chapter, you will learn how derivatives can be used to answer
these questions. You will also see why these questions are important in
real-life applications.

A function need not have a minimum or a maximum on an interval. For
instance, in Figure 3.1(a) and (b), you can see that the function f(x) = x2
+ 1
has both a minimum and a maximum on the closed interval [–1, 2], but
does not have a maximum on the open interval (–1, 2).
Figure 3.1(a) Figure 3.1(b)

Moreover, in Figure 3.1(c),
you can see that continuity
(or the lack of it) can affect the
existence of an extremum on
the interval.
This suggests the theorem below.
Figure 3.1(c)

Relative Extrema and Critical Numbers

In Figure 3.2, the graph of f(x) = x3
– 3x2
has a relative maximum
at the point (0, 0) and a relative minimum at the point (2, –4).
Informally, for a continuous function,
you can think of a relative maximum
as occurring on a “hill” on the graph,
and a relative minimum as occurring
in a “valley” on the graph.
Figure 3.2

Such a hill and valley can occur in two ways.
When the hill (or valley) is smooth and rounded, the graph has a
horizontal tangent line at the high point (or low point).
When the hill (or valley) is sharp and peaked, the graph represents
a function that is not differentiable at the high point (or low point).

Example 1 – The Value of the Derivative at Relative Extrema
Find the value of the derivative at each relative extremum
shown in Figure 3.3.
Figure 3.3

Example 1(a) – Solution
At the point (3, 2), the value of the derivative is f'(3) = 0
[see Figure 3.3(a)].
The derivative of is
Figure 3.3(a)

Example 1(b) – Solution
At x = 0, the derivative of f(x) = |x| does not exist because the following
one-sided limits differ [see Figure 3.3(b)].
cont’d
Figure 3.3(b)

Example 1(c) – Solution
The derivative of f(x) = sin x is f'(x) = cos x.
At the point (π/2, 1), the value of the
derivative is f'(π/2) = cos(π/2) = 0.
At the point (3π/2, –1), the value of the
derivative is f'(3π/2) = cos(3π/2) = 0
[see Figure 3.3(c)].
cont’d
Figure 3.3(c)

Note in Example 1 that at each relative extremum, the derivative either is
zero or does not exist. The x-values at these special points are called
critical numbers.
Figure 3.4 illustrates the two types of critical numbers.
Figure 3.4

Notice in the definition above that the critical number c has
to be in the domain of f, but c does not have to been in the
domain of f'.

Finding Extrema on a Closed Interval
Theorem 3.2 states that the relative extrema of a function
can occur only at the critical number of the function.
Knowing this, you can use the following guidelines to find
extrema on a closed interval.

Example 2 – Finding Extrema on a Closed Interval
Find the extrema of f(x) = 3x4
– 4x3
on the interval [–1, 2].
Solution:
Begin by differentiating the function.
f(x) = 3x4
– 4x3
Write original function.
f'(x) = 12x3
– 12x2
Differentiate.

Example 2 – Solution
To find the critical numbers on the interval (– 1,2), you
must find all x-values for which f'(x) = 0 and all x-values
for which f'(x) does not exist.
f'(x) = 12x3
– 12x2
= 0 Set f'(x) equal to 0.
12x2
(x– 1) = 0 Factor.
x = 0, 1 Critical numbers
Because f' is defined for all x, you can conclude that
these are the only critical numbers of f.
cont’d

By evaluating f at these two critical numbers and at the
endpoints of [–1, 2], you can determine that the
maximum is f(2) = 16 and the minimum is f(1) = –1, as
shown in the table.
Example 2 – Solution cont’d

The graph of f is shown in Figure 3.5.
Figure 3.5
Example 2 – Solution cont’d

cont’d
In Figure 3.5, note that the critical number x = 0 does not
yield a relative minimum or a relative maximum.
This tells you that the converse of Theorem 3.2 is not true.
In other words, the critical numbers of a function need not
produce relative extrema.
Example 2 – Solution

• Determine the absolute extrema for the following function and
interval[-4,2].
•

First, a review problem:
Consider a sphere of radius 10 cm.
If the radius changes 0.1 cm (a very small amount)
how much does the volume change?
34
3
V rπ=
2
4dV r drπ=
( )
2
4 10cm 0.1cmdV π= ⋅
3
40 cmdV π=
The volume would change by approximately 40π cm3
Related Rates

Now, suppose that the radius is
changing at an instantaneous rate
of 0.1 cm/sec.
34
3
V rπ= 2
4
dV dr
r
dt dt
π=
( )
2 cm
4 10cm 0.1
sec
dV
dt
π
 
= ⋅ 
 
3
cm
40
sec
dV
dt
π=
The sphere is growing at a rate of 40π cm3
/sec .
Note: This is an exact answer, not an approximation like
we got with the differential problems.
Related Rates

Water is draining from a cylindrical tank
at 3 liters/second. How fast is the surface
dropping?
L
3
sec
dV
dt
= −
3
cm
3000
sec
= −
Find
dh
dt
2
V r hπ=
2dV dh
r
dt dt
π=
(r is a constant.)
3
2cm
3000
sec
dh
r
dt
π− =
3
2
cm
3000
secdh
dt rπ
= −
(We need a formula to
relate V and h. )
Related Rates

Steps for Related Rates Problems:
1. Draw a picture (sketch).
2. Write down known information.
3. Write down what you are looking for.
4. Write an equation to relate the variables.
5. Differentiate both sides with respect to t.
6. Evaluate.
Related Rates

Hot Air Balloon Problem:
Given:
4
π
θ =
rad
0.14
min
d
dt
θ
=
How fast is the balloon rising?
Find
dh
dt
tan
500
h
θ =
2 1
sec
500
d dh
dt dt
θ
θ =
( )
2
1
sec 0.14
4 500
dh
dt
π 
= 
 
h
θ
500ft
Related Rates

Hot Air Balloon Problem:
Given:
4
π
θ =
rad
0.14
min
d
dt
θ
=
How fast is the balloon rising?
Find
dh
dt
tan
500
h
θ =
2 1
sec
500
d dh
dt dt
θ
θ = ( )
2
1
sec 0.14
4 500
dh
dt
π 
= 
 
h
θ
500ft
( ) ( )
2
2 0.14 500
dh
dt
⋅ =
1
1
2
4
π
sec 2
4
π
=
ft
140
min
dh
dt
=
Related Rates

4x =
3y =
B
A
5z =
Truck Problem:
Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the
trucks changing 6 minutes later?
r t d⋅ =
1
40 4
10
⋅ =
1
30 3
10
⋅ =
2 2 2
3 4 z+ =
2
9 16 z+ = 2
25 z= 5 z=
Related Rates

4x =
3y =
30
dy
dt
=
40
dx
dt
=
B
A
5z =
Truck Problem:
r t d⋅ =
1
40 4
10
⋅ =
1
30 3
10
⋅ =
2 2 2
3 4 z+ =
2
9 16 z+ =
2 2 2
x y z+ =
2 2 2
dx dy dz
x y z
dt dt dt
+ =
4 40 3 30 5
dz
dt
⋅ + ⋅ =
Related Rates

250 5
dz
dt
= 50
dz
dt
=
miles
50
hour
4Related Rates
Truck Problem:

A stone dropped in a pond sends
out a circular ripple whose radius
increases at a constant rate of 4
ft/sec. After 12 seconds, how
rapidly is the area enclosed by the
ripple increasing?

Library Work # 2
•1. Sketch the graph of f. Find the absolute
and local maximum and minimum values of f.
Use the interval [-1,4]
•2. f(x)= X4
+ 4x3
-2x2
-12x

Maxima Minima Problems
•3. A farmer has 800 m fencing material to enclose a
rectangular pen adjacent to a long existing wall. He
will use the wall for one side of the pen and the
available fencing material for the remaining three
sides. What is the maximum area that can be
enclosed this way?

4. Differentiate each of the following:

.
10x4
− 18xy2
+ 10y3
= 48

5. Find the derivative of each function.
5. y = e–3xsin4x

6. Solve the Related Rates problem.
•An airplane is flying west at 500 ft/sec at an altitude
of 4000 ft and a searchlight on the ground lies
directly under the path of the plane. If the light is
kept on the plane , how fast is the searchlight
revolving when the airline distance of the plane
from the searchlight is 2000 feet due east?

Differential calculus

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