Find the derivative of the function f(x) = x2 – 2x
f
f ( x  h)  f ( x )
'( x)  lim
h 0
h
( x  h)2  2( x  h)   ( x 2  2 x)
 lim
h 0
h
x  2 xh  h  2 x  2h  x  2 x
 lim
h 0
h
2
2
2
2 xh  h  2h  lim 2 x  h  2  2x  2
 lim
h 0
h 0
h
2
State Standard – 4.4 Students derive derivative formulas
and use them to find the derivatives of algebraic, trig,
exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
RULE 1
Derivative of a Constant Function
If f has the constant value f(x) = c, then
df
d
 (c )  0
dx dx
If the derivative of a function is its slope, then for a
constant function, the derivative must be zero.
Example 1a
Find the derivative of f(x) = 8
df
dx
d
 (8)
dx
0
Example 1b
 
Find the derivative of f ( x)    
 2
df
d  
  
dx
dx  2 
0
RULE 2
Power Rule for Positive Integers
If n is a positive integer, then
d n
n 1
x  nx
dx
In the Warm-Up we saw that if
y  x  2x
2
,
y  2 x  2
This is part of a pattern.
d n
n 1
x

nx


dx
power rule
examples:
f  x  x
yx
4
f   x   4x
3
8
y  8 x
7
y x
yx
y 
1
 
2
1
y  x
2
1
 1
 
 2
2x
y 
1
 
 2
1
2 x
RULE 3
Constant Mulitple Rule
If u is a differentiable function of x, and c is a constant,
then
d
du
cu  c
dx
dx
constant multiple rule:
examples:
d
du
 cu   c
dx
dx
d n
n 1
cx  cnx
dx
d
5
4
4
7 x  7  5 x  35 x
dx
RULE 4
The Sum Rule
If f and g are both differentiable, then
d
du dv
(u  v) 

dx
dx dx
Example
Find the derivative of
y = x4 + 12x
dy d 4
d
 ( x )  (12 x)
dx dx
dx
 4 x  12
3
RULE 5
The Difference Rule
If f and g are both differntiable, then
d
du dv
(u  v) 

dx
dx dx
Example
Find the derivative of
y = x3 – 3x
dy d 3
d
 ( x )  (3 x)
dx dx
dx
 3x  3
2
RULE 6 The Derivative of the Natural Exponential Function
d x
x
(e )  e
dx
Pg. 191
3 – 31 odd
Find the derivative of the function
2
f(x) = 3x – 5x + 1
 6x  5
State Standard – 4.4 Students derive derivative formulas
and use them to find the derivatives of algebraic, trig,
exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
Derivative of Sine and Cosine Functions:
d
sin x   cos x
dx
d
 cos x    sin x
dx
THE PRODUCT RULE:
d
 f ( x) g ( x)  f ( x) g ( x)  g ( x) f ( x)
dx
Example 1
f(x)
g(x)
Find the derivative of f(x) = (2x+5)(3+4x)
f ( x)  f ( x) g ( x)  g ( x) f ( x)
f ( x)  (2 x  5)(4)  (3  4 x)(2)
f ( x)  8 x  20  6  8 x
f ( x)  16 x  26
Example 2
f(x)
g(x)
Find the derivative of f(x) = (4x3)(sin x)
f ( x)  f ( x) g ( x)  g ( x) f ( x)
3
2

f ( x)  (4 x )(cos x)  (sin x)(12 x )
f ( x)  4 x cos x  12 x sin x
3
2
Example 3
Find the derivative of y = (5x2)(cos x) + (3x)(sinx)
y  f ( x) g ( x)  g ( x) f ( x)
y  (5x 2 )( sin x)  (cos x)(10 x)  (3x)(cos x)  (sin x)(3)
y  5x 2 sin x  10 x cos x  3x cos x  3sin x
y  5x 2 sin x  13x cos x  3sin x
3-25 odd
Find the derivative of the function
f(x) = (x – 4)(x + 3)
f ( x)  f ( x) g ( x)  g ( x) f ( x)
f ( x)  ( x  4)(1)  ( x  3)(1)
f ( x)  x  4  x  3
f ( x)  2 x  1
State Standard – 4.4 Students derive derivative formulas
and use them to find the derivatives of algebraic, trig,
exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
THE QUOTIENT RULE:
d  f ( x)  g ( x) f ( x)  f ( x) g ( x)



2
dx  g ( x) 
 g ( x)
Example 1
Find the derivative of
f ( x) 
f(x)
2x 1
f ( x) 
x
g(x)
g ( x) f ( x)  f ( x) g ( x)
 g ( x) 
2
( x)(2)  (2 x  1)(1)
f ( x) 
2
x
2x  2x 1
f ( x) 
2
x
1
f ( x)   2
x
Example 2
Find the derivative of
f ( x) 
f(x)
3x  4
f ( x) 
5x  2
g(x)
g ( x) f ( x)  f ( x) g ( x)
 g ( x) 
2
(5 x  2)(3)  (3x  4)(5)
f ( x) 
2
(5 x  2)
15 x  6  15 x  20
f ( x) 
(5 x  2) 2
26
f ( x) 
2
(5 x  2)
Example 3
Find the derivative of
f ( x) 
f(x)
cos x
f ( x)  3
x
g(x)
g ( x) f ( x)  f ( x) g ( x)
 g ( x) 
2
( x )( sin x)  (cos x)(3x )
f ( x) 
3 2
(x )
3
x ( x sin x  3cos x)
f ( x) 
6
x
2
2
 x sin x  3cos x
f ( x) 
x4
3 – 25 odd
State Standard – 4.4 Students derive derivative formulas
and use them to find the derivatives of algebraic, trig,
exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
Derivatives of the remaining trig functions:
d
sin x  cos x
dx
d
cot x   csc 2 x
dx
d
cos x   sin x
dx
d
sec x  sec x  tan x
dx
d
tan x  sec 2 x
dx
d
csc x   csc x  cot x
dx
Example 1
Find the derivative of y = (sec x)(tan x)
y  f ( x) g ( x)  g ( x) f ( x)
y  (sec x)(sec x)  (tan x)(sec x tan x)
2
y  sec x  sec x tan x
3
2
y  sec x  sec x(sec x  1)
3
2
y  sec x  sec x  sec x
3
3
3

y  2sec x  sec x
Higher Order Derivatives:
y 
dy
dx
is the first derivative of y with respect to x.
dy d dy d 2 y
y 

 2
dx dx dx dx
dy
y 
dx
y
 4
is the second derivative.
(y double prime)
is the third derivative.
d

y is the fourth derivative.
dx
We will learn
later what these
higher order
derivatives are
used for.

y  10 x  33x  15
4
2
y  40 x  66 x
'
3
y  120 x  66
''
2
y  240 x
'''
y
 4
 240
WS
Find the Derivative:
d
1) sin x  cos x
dx
d
4) cot x   csc 2 x
dx
d
2) cos x   sin x
dx
d
5) sec x  sec x  tan x
dx
d
3) tan x  sec 2 x
dx
d
6) csc x   csc x  cot x
dx
State Standard – 4.4 Students derive derivative formulas
and use them to find the derivatives of algebraic, trig,
exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
So far we have been memorizing the derivatives of the trig
functions. And today we will be investigating this further.
This will help us as we go into the next section of using the
Chain Rule.
2) Find the derivative of f(x) = (x)(sin x)
f ( x)  f ( x) g ( x)  g ( x) f ( x)
f ( x)  ( x)(cos x)  (sin x)(1)
f ( x)  x cos x  sin x
ON WHITE BOARD
a) Find the derivative of f(x) = (x)(cos x)
f ( x)  f ( x) g ( x)  g ( x) f ( x)
f ( x)  ( x)( sin x)  (cos x)(1)
f ( x)   x sin x  cos x
b) Find the derivative of f(x) = (x)(tan x)
f ( x)  f ( x) g ( x)  g ( x) f ( x)
2

f ( x)  ( x)(sec x)  (tan x)(1)
f ( x)  x sec x  tan x
2
4) Find the derivative of y = 2 csc x + 5 cos x
y  2( csc x cot x)  5( sin x)
y  2 csc x cot x  5sin x
ON WHITE BOARD
a) Find the derivative of y = 4 sec x + 3 sin x
y  4(sec x tan x)  3(cos x)
y  4sec x tan x  3cos x
b) Find the derivative of y = 7 cot x + 2 tan x
2
2

y  7( csc x)  2(sec x)
y  7 csc x  2sec x
2
2
1  sin x
y
x  cos x
10) Find the derivative of
f ( x) 
g ( x) f ( x)  f ( x) g ( x)
 g ( x) 
2
( x  cos x)(cos x)  (1  sin x)(1  sin x)
y 
2
( x  cos x)
x cos x  cos x  (1  sin x)
y 
2
( x  cos x)
1
2
2
x cos x  cos x  1  sin x
y 
2
( x  cos x)
2
2
x cos x
y 
2
( x  cos x)
ON WHITE BOARD
12)
tan x  1
y
sec x
f ( x) 
g ( x) f ( x)  f ( x) g ( x)
 g ( x) 
2
(sec x)(sec x)  (tan x  1)(sec x tan x)
y 
2
(sec x)
2
sec x  (sec x tan x  sec x tan x)
y 
2
sec x
3
2
sec x  sec x tan x  sec x tan x
y 
2
sec x
3
sec2 x  tan 2 x  tan x
y 
sec x
2
1  tan 2 x  sec 2 x
y 
1  tan x
sec x
Pg. 216
1 – 15 odd
Find y´´
y  csc x
y   csc x cot x
y   (csc x)( csc 2 x)  (cot x)( csc x cot x) 
y     csc3 x  csc x cot 2 x) 
y  csc3 x  csc x cot 2 x
y  csc x(csc2 x  cot 2 x)
y  csc x(csc2 x  csc2 x  1)
y  csc(2csc2 x  1)
y  2csc3 x  csc x
State Standard – 5.0 Students know the Chain Rule and its
proof and applications to the calculation of the derivative
of a variety of composite functions.
Objective – To be able to use the Chain Rule to solve
applications.
We now have a pretty good list of “shortcuts” to
find derivatives of simple functions.
Of course, many of the functions that we will
encounter are not so simple. What is needed is a
way to combine derivative rules to evaluate
more complicated functions.
Consider a simple composite function:
y  6 x  10
y  2  3x  5 
If u  3x  5
then y  2u
y  6 x  10 y  2u
dy
dy
2
6
du
dx
6  23
dy dy du


dx du dx
u  3x  5
du
3
dx
one more:
y  9x  6x 1
2
y   3 x  1
y u
2
u  3x  1
2
If u  3x  1
then y  u 2
y  9x2  6x  1
du
dy
dy
 18 x  6
3
 2u
dx
du
dx
dy
 2  3 x  1
du
dy
 6x  2
du
This pattern is called
the chain rule.
18x  6   6 x  2   3
dy dy du


dx du dx
The Chain Rule can be written either in
Leibniz notation:
dy dy du


dx du dx
Or in Prime Notation:
F ( x)  f ( g ( x)) g ( x)
Example 1
dy
Find
of
dx
u  4  3x
y u
9
y  (4  3x)
du
 3
dx
dy
 9u 8
du
9
dy du

 9u 8 ( 3)
du dx
 27u
8
dy
8
 27(4  3x)
dx
On White Board
dy
Find
of
dx
u  5x  x
3
y u
7
4
y  (5x  x )
3
4 7
du
 15 x 2  4 x3
dx
dy
 7u 6
du
dy du
2
3
6

 7u (15 x  4 x )
du dx
dy
3
4 6
2
3
 7(5 x  x ) (15 x  4 x )
dx
On White Board
dy
Find
of
dx
y  (3x  7 x  5)
u  3x  7 x  5
4
yu
3
4
3
du
 12 x 3  7
dx
dy
2
 3u
du
dy du
3
2

 3u (12 x  7)
du dx
dy
4
2
3
 3(3x  7 x  5) (12 x  7)
dx
9 – 42 mult of 3
Find y´´
y  sec x
y  sec x tan x
y  sec x(sec2 x)  tan x(sec x tan x)
2
2

y  sec x(sec x  tan x)
y  sec x(sec2 x  sec2 x  1)
y  sec x(2sec2 x  1)
y  2sec3 x  sec x
State Standard – 5.0 Students know the Chain Rule and its
proof and applications to the calculation of the derivative
of a variety of composite functions.
Objective – To be able to use the Chain Rule to solve
applications.
The Chain Rule can be written either in
Leibniz notation:
dy dy du


dx du dx
Or in Prime Notation:
F ( x)  f ( g ( x)) g ( x)
Example 1
dy
Find
of
dx
u  sin x  3
y u
9
y  (sin x  3)
du
 cos x
dx
dy
 9u 8
du
9
dy du

 9u 8 (cos x )
du dx
 9cos x(u )
8
dy
8
 9 cos x(sin x  3)
dx
On White Board
dy
Find
of
dx
3 x
u
2
y  sin u  cos u
 3 x 
 3 x 
y  sin 
  cos 

 2 
 2 
du 3

dx
2
dy
 cos u  sin u
du
dy du (cos u  sin u ) 3



2
du dx
dy 3   3 x 
 3 x  

cos 
  sin 


dx 2   2 
 2 
On White Board
dy
Find
of
dx
u  x 1
2
y  cos u
y  cos( x  1)
2
du
 2x
dx
dy
  sin u
du
dy du


du dx
 sinu (2 x)
dy
2
 2 x sin( x  1)
dx
9 – 42 mult of 3
Find the Derivative
1)sin( x  2 x)
8)cos( x3  4 x)
15)3(tan 4 x)
2)tan 4x
9)sin 5x
16) sin 5x
2
2
2
3)sin(3x )
10)cos( x3 ) 2
4)cos10x
11)cos 4x
3 2
2
12)tan x 2
5)cos x 2
6)cos x)
2
2
7)sin(3x3 )
13)sin x)
2
14)cos( x5 )
17)  cos(2 x3 )
Find y´´
y  sec x
y  sec x tan x
y  sec x(sec2 x)  tan x(sec x tan x)
2
2

y  sec x(sec x  tan x)
y  sec x(sec2 x  sec2 x  1)
y  sec x(2sec2 x  1)
y  2sec3 x  sec x
State Standard – 5.0 Students know the Chain Rule and its
proof and applications to the calculation of the derivative
of a variety of composite functions.
Objective – To be able to use the Chain Rule to solve
applications.
Repeated Use of the Chain Rule
We sometimes have to use the Chain Rule two
or more times to find a derivative.
Example 1
Find the derivative of g(t) = tan (5 – sin 2t)
d
g (t )  (tan(5  sin 2t ))
dt
u  5  sin 2t
y  tan u
du d
 (5  sin 2t )
dt dt
dy
 sec 2 u
du
dy du
d
2

 sec u  (5  sin 2t )
dt
du dt
 sec (5  sin 2t ) 
2
u  2t
y  5  sin u
du
2
dt
d
(5  sin 2t )
dt
dy
  cos u
du
2

g (t )  sec (5  sin 2t ) (2cos 2t )
Example 2
Find the derivative of f(x) = cos ( sin 3x)
d
f ( x)  (cos(sin 3t ))
dx
u  sin3x
y  cos u
du d
 (sin 3 x)
dx dx
dy
  sin u
du
dy du
d

sinu


 (sin 3 x)
dx
du dx
  sin(sin 3 x) 
u  3x
y  sin u
du
3
dx
d
(sin 3 x)
dx
dy
 cos u
du
f ( x)   sin(sin 3x) (3cos 3 x)
Find the Derivative
1) y  sin(cos(2 x  5))

 x 
2) y  1  tan   
 12  

3
3) y  1  cos( x )
2
9 – 42 mult of 3
Find the Derivative
1) y  tan(sin(3 x  5)) 6) y  1  tan( x 2 )
2
2) y  sin(cos( x ))
 x
7) y  1  cos  
 10 
3) y  cos (3 x  2 x)
2
2

 x 
4) y  3  sec   
 3 

8) y  sin ( x  4) cos ( x )
3
5) y   4 x  cos  3 x  
2
3
2
9) y  tan (3 x  1)
4
2
10) y  cos (cos (cos x))
3
2
2
Find
Find y´´
the Derivative:
d
1) sin x  cos x
dx
d
4) cot x   csc 2 x
dx
d
2) cos x   sin x
dx
d
5) sec x  sec x  tan x
dx
d
3) tan x  sec 2 x
dx
d
6) csc x   csc x  cot x
dx
State Standard – 6.0 Students use implicit differentiation in
a wide variety of problems.
Objective – To be able to use implicit differentiation
Implicit Differentiation
(Takes Four Steps)
1) Differentiate both sides of the equation with respect
to ‘x’, treating ‘y’ as a differentiable function of ‘x’.
2) Collect the terms with the dy/dx on one side of the
equation.
3) Factor out the dy/dx.
4) Solve for
dy/
dx.
x  y 1
2
2
d 2 d 2 d
x 
y  1
dx
dx
dx
This is not a function,
but it would still be
nice to be able to find
the slope.
Do the same thing to both sides.
Note use of chain rule.
dy
2x  2 y
0
dx
dy
2y
 2 x
dx
dy 2 x

dx 2 y
dy
x

dx
y
2 y  x  sin y
2
This can’t be solved for y.
d
d 2 d
2y 
x  sin y
dy
2x
dx
dx
dx

dx 2  cos y
dy
dy
2  2 x  cos y
dx
dx
dy
dy
2  cos y
 2x
dx
dx
This technique is called
implicit differentiation.
1 Differentiate both sides w.r.t. x.
dy
 2  cos y   2 x
dx
dy
2 Solve for
.
dx
Example 1
dy
Find
of
dx
2x  3 y  4
2
2
4 x  6 y ( y)  0
4x
4x
6 y ( y)  4 x
6 y
2x
y 
3y
6 y
Example 2
dy
Find
of
dx
x  y  5x  4 y
2
2
2 x  2 y ( y)  5  4( y)
2x
4( y)
2x
4( y)
2 y ( y)  4( y)  5  2 x
y(2 y  4)  5  2 x
2y  4
2y  4
5  2x
y 
2y  4
Example 3
dy
Find
of
dx
x  y  3xy
3
3
3x  3 y ( y)  3 x(1)( y)  (1)( y)
2
2
3
3
x  y ( y)  x( y)  y
2
2
 x ( y )
 x2
 x ( y )
x
y ( y)  x( y)   x  y
2
2
y( y  x)   x  y
2
2
2
y x
2
y2  x
x  y
y  2
y x
2
Example 4
dy
Find
of
dx
2sin x cos y  1
2 (sin x)( sin y)( y)  (cos y)(cos x)  0
2
2
(sin x)( sin y )( y)  (cos y)(cos x)  0
(cos y )(cos x)
(cos y )(cos x)
(sin x)( sin y )( y)  (cos y )(cos x)
(sin x)( sin y )
(sin x)( sin y )
cos y cos x
y 
sin x sin y
Pg. 233
1a, 2a, 3a, 5 – 9, 11,
13, 14, and 20
From www.Dictionary.com
Explicit6.Mathematics. (of a function) having the dependent
variable expressed directly in terms of the
independent variables, as y = 3x + 4.
Implicit4.Mathematics. (of a function) having the dependent
variable not explicitly expressed in terms of the
independent variables, as x2 + y2 = 1.
dy
Find
of
dx
3x  y  2 x  5 y
2
2
6 x  2 y ( y)  2  5( y)
6x
5( y)
6x
5( y)
2 y ( y)  5( y)  2  6 x
y(2 y  5)  2  6 x
2y 5
2y 5
2  6x
y 
2y 5
State Standard – 4.4 Students derive derivative formulas
and use them to find the derivatives of inverse trig
functions.
Objective – To be able to take derivatives of Inverse Trig
Functions.
1
sin x  arcsin x
d
1
1
(sin u ) 
u
dx
1 u2
d
1
1
(csc u ) 
u
dx
u u2 1
d
1
1
(cos u ) 
u
dx
1 u2
d
1
1
(sec u ) 
u
dx
u u2 1
d
1
1
(tan u ) 
u
2
dx
1 u
d
1
1
(cot u ) 
u
2
dx
1 u
d
1
1
(sin u ) 
u
dx
1 u2
1
sin x  y
d
dx
x  sin y
1
d
dx
dy
1  cos y
dx
dy
1

dx cos y
1
x
b2  1  x2
y
b  1  x2
b
adj
1  x2

hyp
1
x 2  b2  1
dy
1

2
dx
1 x
Example 1
Find the derivative of f ( x)  sin 1 x 2
u  x2
d
1
1
(sin u ) 
u
dx
1 u2
f ( x) 
1
1 (x )
f ( x) 
2 2
2x
1 x
4
2x
Example 2
Find the derivative of f ( x)  tan 1
1
f ( x)  tan x
f ( x) 
1
 
1  x 
 
1
2
2
1
x
2
1
1
f ( x) 
1 x 2 x
1
f ( x) 
(2)(1  x) x
1

2
 
x
1
2
ux
1
2
d
1
1
(tan u ) 
u
2
dx
1 u
1
f ( x) 
(2 x  2) x
Find the Derivative.
1
1) y  cos ( x )
2
1
2) y  cos  
x
1
1
3) y  sin 2 x
4) y  sin
1
1  x 
1
5) y  sec (2s  1)
6) y  tan
1
3x 1
x
7) y  csc
2
1 3
8) y  sin 2
x
1
In Sec. 3.2 we learned the basics of Higher Order Derivatives.
So Find the first four derivatives of y  x3  3x 2  2
First Derivative:
y  3 x  6 x
Second Derivative:
y  6 x  6
Third Derivative:
y  6
Fourth Derivative:
2
y
(4)
0
Velocity and Acceleration
State Standard – 7.0 Students compute derivatives of
higher order.
Objective – To be able to solve problems involving
multiple derivative steps.
Higher Order Derivatives:
y 
dy
dx
is the first derivative of y with respect to x.
dy d dy d 2 y
y 

 2
dx dx dx dx
dy
y 
dx
y
 4
is the second derivative.
(y double prime)
is the third derivative.
d

y is the fourth derivative.
dx
We will learn
later what these
higher order
derivatives are
used for.
Example 1
Find the second derivative of
y  2 x sin x
y  (2 x)(cos x)  (sin x)(2)
y  2 x cos x  2sin x
y  (2 x)( sin x)  (cos x)(2)  2cos x
y  2 x sin x  2 cos x  2 cos x
y  2 x sin x  4 cos x
Example 2
Find the second derivative of
f ( x) g ( x)  f ( x) g ( x)
y 
[ g ( x)]2
(1)( x 2 )  ( x  1)(2 x)
y 
( x 2 )2
x  2x  2x
y 
4
x
2
2
x 1
y 2
x
(1)( x3 )  ( x  2)(3x 2 )
y 
( x3 ) 2
 x3  (3x3  6 x 2 )
y 
x6
 x3  3x3  6 x 2
y 
x6
3
2
x  2
x  2x
2
x

6
x
2
x

6

y 

y 
3
4

y

6
x
x
x
x4
2
s  s(t ) is the position function
Velocity is the first derivative of position.
ds
v(t )  s(t ) 
dt
Acceleration is the first derivative of Velocity.
a(t )  v(t )  s(t )
Acceleration is the second derivative of position.
Example 3
The position of a particle is given by the equation:
s  f (t )  t  6t  9t
3
2
Where t is measured in seconds and s in meters.
a) Find the acceleration at time t. What is the acceleration after 4
seconds.
ds
v(t ) 
 3t 2  12t  9
dt
dv
a(t ) 
 6t  12
dt
a(4)  6(4)  12  24  12
m
a (4)  12 2
s
Pg. 240
5 – 15 odd, 29, 31, 43 and
44
The position of a particle is given by the equation:
s  f (t )  2t  5t  4t
3
2
Where t is measured in seconds and s in meters.
a) Find the acceleration at time t. What is the acceleration after 5
seconds.
ds
v(t ) 
 6t 2  10t  4
dt
dv
a (t ) 
 12t  10
dt
a(5)  12(5)  10  60 10
m
a (5)  50 2
s
A dynamite blast blows a heavy rock straight up with a
launch velocity of 160 ft/sec. It reaches a height of:
s  160t  16t ft after t sec.
2
Where t is measured in seconds and s in feet.
a) How high does the rock go?
ds
v(t ) 
 160  32t ft / sec
dt
What is the velocity at the rock’s height?
0  160  32t
32t  160
t  5sec
s(5)  160(5)  16(5)2
s (5)  800  400
s (5)  400 ft
State Standard – 4.4 Students find the derivatives of
logarithmic functions.
Objective – Students will be able to solve problems
involving logarithmic functions.
Derivative of Log Functions
d
1
ln x  
dx
x
d
1
ln u   u
dx
u
Example 1
a ) y  ln 8 x
b) y  ln( x 2  2)
1
y 
8
8x
1
y  2
2x
x 2
1
y 
x
2x
y  2
x 2
Derivative of Log Functions
d
1
ln x  
dx
x
d
1
ln u   u
dx
u
Example 2
a ) y  x ln x
1

y  ( x)    (ln x)(1)
x
y  1  ln x
b) y  (ln x)
y  3(ln x)
2
3
1
x
3(ln x) 2
y 
x
More Derivatives
d x
e   e x
dx
d u
e   eu u 
dx
Example 3
ye
2x
2x

y  (e )  2 
y  2e
2x
Derivative of Log Functions (Base other than e)
d
 a x   (ln a )(a x )
dx
d u
 a   (ln a )a u u 
dx
Example 4
a) y  2
b) y  2
3x
x
y  (ln 2)(2 )
y  (ln 2)23 x 3
y  2 (ln 2)
3x

y  (3ln 2)(2 )
x
x
Derivative of Log Functions (Base other than e)
d
1
log a x  
dx
(ln a) x
d
1
u
log a u  
dx
(ln a)u
Example 5
a) y  log 2 x
1
y 
(ln 2) x
b) y  log 3 x
4
1
3
y 
4
x
(ln 3) x 4
4
y 
(ln 3) x
Pg. 249
2 – 5, 7, 9, 21 – 23, and 30
Find the First Derivative:
1)
y6
x
x

y  6 ln 6
Find the second Derivative:
2)
y  cos 2 x
y  ( sin 2 x)(2)
y  2sin 2 x
y  (2)(cos 2 x)(2)
y  4 cos 2 x
State Standard – 4.2 Students demonstrate an
understanding of the interpretation of the derivative as an
instantaneous rate of change.
Objective – Students will be able to solve problems
involving rate of change.
If we are pumping air into a balloon, both the
volume and the radius of the balloon are
increasing and their rates of increase are
related to each other. We will find an equation
that relates the two quantities and then use the
Chain Rule to differentiate both sides with
respect to time.
Example 1
Air is being pumped into a spherical
balloon so that its volume increases
at a rate of 100 cm3/s. How fast is the
radius of the balloon increasing when the
diameter is 50 cm?
dr
1 dv
dv
cm3

Given :
 100
2
dt
4

r
dt
dt
s
dr
Unknown :
when r  25 cm
dt
4 3
V  r
3
dv 4
2 dr
  (3r )
dt 3
dt
dv
2 dr
 4 r
dt
dt
2
2
4 r 4 r

dr
1
cm3 

100

2 
dt 4 (25) 
s 
dr
1 cm

dt 25 s
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.
Example 2
How rapidly will the fluid level
inside a vertical cylindrical tank
drop if we pump the fluid out at
the rate of 3000 L/min?
dh
?
dt
dv
L
 3000
dt
min
V   r 2h
But since there are 1000L
in a cubic meter.
V  1000 r h
2
dv
2 dh
 1000 (r )
dt
dt
h
dh
3000  1000 r
dt
2
1000 r 1000 r
2
dh
3000

dt 1000 r 2
2
dh 3
 2
dt  r
The fluid level
will drop at the
rate of 3 2 m
 r min
Example 3
2
A water tank has the shape of an
inverted circular cone with base
radius 2m and height 4m. If water
is being pumped into the tank at a
rate of 2 m3/min, find the rate of
rising water when the water is 3 m
deep.
1 2
V  r h
3
r 2
h

r
2
h 4
V

12
r
h3
dv 
2 dh
 (3h )
dt 12
dt
2
1 h
V    h
3 2
4
dh
?
dt
2
dv

h
dh 4
4

2
2 dt

h
4
dt
h
h
dv
m3
2
dt
min
dh
4 dv
 2
dt  h dt
dh
4  m3 

2

2 
dt  (3)  min 
dh 8 m
m

 0.28
dt 9 min
min
Example 4
5
Water runs into a conical tank at
the rate of 9 ft3/min. The tank
stands point down and has a height
of 10 ft. and a base radius of 5 ft.
How fast is the water level rising
when the water is 6 ft. deep?
1 2
V  r h
3
r 5
h

r
h 10
2
V

12
r
h3
dv 
2 dh
 (3h )
dt 12
dt
2
1 h
V    h
3 2
10
dh
?
dt
2
dv

h
dh 4
4

2
2 dt

h
4
dt
h
h
dv
ft 3
9
dt
min
dh
4 dv
 2
dt  h dt
dh
4  ft 3 

9

2 
dt  (6)  min 
dh 1 ft
ft

 0.32
dt  min
min
Pg. 260
1, 3 – 5, 7, 10, and 19