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3.1 Derivative of a Function
lim
h 0
f a  h  f a
h
We write:
f
is called the derivative of
f   x   lim
h 0
at
a.
f a  h  f a
h
“The derivative of f with respect to x is …”
There are many ways to write the derivative of
y  f  x
3.1 Derivative of a Function
f  x
“f prime x”
y
“y prime”
or
“the derivative of f with respect
to x”
dy
dx
“dee why dee ecks”
or
“the derivative of y with
respect to x”
df
dx
“dee eff dee ecks”
or
“the derivative of f with
respect to x”
d
f
dx
 x  “dee dee ecks uv eff uv ecks” or “the derivative
( d dx of f of x )
of f of x”

3.1 Derivative of a Function
dx does not mean d times x !
dy does not mean d times y !
3.1 Derivative of a Function
dy
does not mean dy  dx !
dx
(except when it is convenient to think of it as division.)
df
does not mean df  dx !
dx
(except when it is convenient to think of it as division.)
3.1 Derivative of a Function
d
d
f  x  does not mean
times f  x  !
dx
dx
(except when it is convenient to treat it that way.)
3.1 Derivative of a Function
4
3
The derivative is
the slope of the
original function.
2
y  f  x
1
0
3
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
2
The derivative is defined
at the end points of a
function on a closed
interval.
1
0
-1
-2
y  f  x
7
8
9
3.1 Derivative of a Function
6
y  x 3
5
2
4
3
2
1
-3
-2
-1
0
-1
1
x
2
3
y  lim
-2
-3
h 0
6
5
4
3
2
1
-3 -2 -1 0
-1
-2
-3
-4
-5
-6
 x  h
y  lim 2 x  h
h 0
1
x
2
3
y  2 x
2

 3  x2  3
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.
3.2 Differentiability
To be differentiable, a function must be continuous
and smooth.
Derivatives will fail to exist at:
f
 x 
x
f  x  x
corner
f
 x 
3
cusp
f
x
vertical tangent
2
3
1, x  0
 1, x  0
 x  
discontinuity
3.2 Differentiability
Most of the functions we study in calculus will be differentiable.
3.2 Differentiability
There are two theorems on page 110:
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
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  .
1
f a 
2
f  b  3
Between a and b, f  must take
1
on every value between 2 and
3.
3.3 Rules for Differentiation
If the derivative of a function is its slope, then for a
constant function, the derivative must be zero.
d
c  0
dx
example:
y3
y  0
The derivative of a constant is zero.
3.3 Rules for Differentiation
We saw that if
y  x 2 , y  2 x
.
This is part of a pattern.
d n
n 1
x

nx


dx
examples:
f  x  x
f   x   4x
power rule
y  x8
4
3
y  8 x 7
3.3 Rules for Differentiation
d n
n 1
x

nx


dx
Proof:
d n
( x  h) n  x n
x  lim
h0
dx
h
d n
x n  nx n1h  ...  hn  x n
x  lim
h0
dx
h
d n
nx n1h  ...  hn
x  lim
h0
dx
h
d n
x  lim nx n1
h 0
dx
3.3 Rules for Differentiation
constant multiple rule:
d
du
 cu   c
dx
dx
examples:
d n
cx  cnx n 1
dx
d
7 x 5  7  5 x 4  35 x 4
dx
3.3 Rules for Differentiation
constant multiple rule:
d
du
 cu   c
dx
dx
sum and difference rules:
d
du dv
u  v   
dx
dx dx
d
du dv
u  v   
dx
dx dx
4
2
y

x

2
x
2
y  x  12 x
(Each term
is treated separately)
dy
3
3
 4x  4x
y  4 x  12
dx
4
3.3 Rules for Differentiation
Find the horizontal tangents of:
y  x  2x  2
4
2
dy
 4 x3  4 x
dx
Horizontal tangents occur when slope = zero.
4 x3  4 x  0
Substituting the x values into the
x3  x  0
x  x  1  0
2
x  x  1 x  1  0
x  0, 1, 1
original equation, we get:
y  2, y  1, y  1
(The function is even, so we
only get two horizontal
tangents.)
3.3 Rules for Differentiation
4
y  x4  2x2  2
3
-2
-1
2
y2
1
y 1
0
-1
-2
1
2
3.3 Rules for Differentiation
4
y  x4  2x2  2
3
2
dy
 4 x3  4 x
dx
1
-2
-1
0
-1
First derivative
(slope) is zero at:
x  0, 1, 1
-2
1
2
3.3 Rules for Differentiation
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

d  2
x 3

dx
 2 x
3

 x 2  3  6 x 2  5    2 x 3  5 x   2x 
 5x 


d
2 x 5  5 x 3  6 x 3  15 x
dx

d
2 x 5  11x 3  15 x
dx

10 x 4  33 x 2  15

6 x 4  5 x 2  18 x 2  15  4 x 4  10 x 2
10 x 4  33 x 2  15
3.3 Rules for Differentiation
product rule:
d
dv
du
uv

u

v
 
dx
dx
dx
d
u ( x  h )v ( x  h )  u ( x )v ( x )
(uv)  lim
h0
dx
h
add and subtract u(x+h)v(x)
in the denominator
d
u ( x  h )v ( x  h )  u ( x )v ( x )  u ( x  h )v ( x )  u ( x  h )v ( x )
(uv)  lim
h0
dx
h
Proof
d
 u ( x  h)v( x  h)  v( x)   v( x)u ( x  h)  u ( x)  
(uv)  lim 

h

0
dx
h


d
dv
du
(uv)  u  v
dx
dx
dx
3.3 Rules for Differentiation
quotient rule:
du
dv
v u
d u
dx
dx

 
dx  v 
v2
d 2 x  5x

2
dx x  3
3
 u  v du  u dv
d 
2
v
v
 
or
2
2
3
x

3
6
x

5

2
x


   5x   2 x 
x
2
 3
2
3.3 Rules for Differentiation
Higher Order Derivatives:
dy
y 
dx
is the first derivative of y with respect to x.
dy
d dy d 2 y
is the second derivative.
y 


dx dx dx dx 2
(y double prime)
dy
y 
is the third derivative.
dx
We will learn
later what these
d
 4
y 
y is the fourth derivative.
higher order
dx
derivatives are
used for.
3.3 Rules for Differentiation
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 :
d
1. (uv)
dx
d
(uv)  uv' vu'
dx
d u
2.  
dx  v 
d  u  vu'uv '
 
dx  v 
v2
(1)( 7)  (5)( 4)
 27
2
1
d v
3.  
dx  u 
d  v  uv 'vu'
 
dx  u 
u2
(5)( 4)  (1)( 7)
52
5(3)  (1)( 7)  8
27

25
3.3 Rules for Differentiation
d  ho 
 
dx  hi 
(hi)d (ho)  (ho)d (hi )
(ho)( ho)
3.3 Rules for Differentiation
3.4 Velocity and other Rates
of Change
Consider a graph of displacement (distance traveled) vs. time.
distance
(miles)
B
A
Average velocity can be found by
taking:
change in position
s

s
change in time
t
t
time (hours)
Vave
f  t  t   f  t 
s


t
t
The speedometer in your car does not measure average
velocity, but instantaneous velocity.
V t  
f  t  t   f  t 
ds
 lim
t 0
dt
t
(The velocity at one
moment in time.)
3.4 Velocity and other Rates
of Change
Velocity is the first derivative of position.
Acceleration is the second derivative
of position.
3.4 Velocity and other Rates
of Change
Example:
Free Fall Equation
1 2
s g t
2
s  16 t
2
1
s   32 t 2
2
ds
V
 32 t
dt
Speed is the absolute value of velocity.
Gravitational
Constants:
ft
g  32
sec 2
m
g  9.8
sec 2
cm
g  980
sec 2
3.4 Velocity and other Rates
of Change
Acceleration is the derivative of velocity.
dv
a
dt
d 2s
 2
dt
If distance is in:
example:
v  32t
a  32
feet
Velocity would be in:
feet
sec
Acceleration would be in:
ft
sec
sec
ft

sec 2
3.4 Velocity and other Rates
of Change
acc neg
vel pos &
decreasing
acc neg
vel neg &
decreasing
acc zero
vel pos &
constant
distance
velocity
zero
acc pos
vel pos &
increasing
acc zero
vel neg &
constant
acc pos
vel neg &
increasing
acc zero,
velocity zero
time
3.4 Velocity and other Rates
of Change
Rates of Change:
Average rate of change =
f  x  h  f  x
h
Instantaneous rate of change = f   x   lim
h 0
These definitions are true for any function.
( x does not have to represent time. )
f  x  h  f  x
h
3.4 Velocity and other Rates
of Change
For a circle:
A   r2
dA d
  r2
dr dr
dA
 2 r
dr
dA  2 r dr
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.
3.4 Velocity and other Rates
of Change
from Economics:
Marginal cost is the first derivative of the cost function, and
represents an approximation of the cost of producing one
more unit.
3.4 Velocity and other Rates
of Change
Example 13:
3
2
c
x

x

6
x
 15x
Suppose it costs:  
2
to produce x stoves.c  x   3x  12 x  15
If you are currently producing 10 stoves,
the 11th stove will cost approximately:
The actual cost is: C 11  C 10
c 10  3 102  12 10  15
 300 120  15
 $195
 113  6 112  15 11  103  6 10 2  15 10 
 770  550  $220
actual cost
marginal cost
3.4 Velocity and other Rates
of Change
Note that this is not a
great approximation –
Don’t let that bother you.
Marginal cost is a linear approximation of a curved
function. For large values it gives a good approximation
of the cost of producing the next item.
3.4 Velocity and other Rates
of Change
3.5 Derivatives of
Trigonometric Functions
Consider the function y  sin  
We could make a graph of the slope:



Now we connect the dots!
The resulting curve is a cosine curve.
d
sin  x   cos x
dx

2
slope
1
0
0

2
1

1
0
3.5 Derivatives of
Trigonometric Functions
Proof
d
sin( x  h)  sin x
sin x  lim
h0
dx
h
d
sin x cos h  sin h cos x  sin x
sin x  lim
h0
dx
h
d
sin x (cos h  1)  sin h cos x
sin x  lim
h0
dx
h
d
sin x (cos h  1)
sin h cos x
sin x  lim
 lim
h0
h0
dx
h
h
3.5 Derivatives of
Trigonometric Functions
=0
=1
d
sin x (cos h  1)
sin h cos x
sin x  lim
 lim
h0
h0
dx
h
h
d
sin  x   cos x
dx
3.5 Derivatives of
Trigonometric Functions
Find the derivative of cos x
d
cos( x  h)  cos x
cos x  lim
h0
dx
h
d
cos x cos h  sin h sin x  cos x
cos x  lim
h0
dx
h
d
cos x (cos h  1)  sin h sin x
cos x  lim
h0
dx
h
d
cos x (cos h  1)
sin h sin x
cos x  lim
 lim
h0
h0
dx
h
h
3.5 Derivatives of
Trigonometric Functions
=0
=1
d
cos x (cos h  1)
sin h sin x
cos x  lim
 lim
h0
h0
dx
h
h
d
cos  x    sin x
dx
3.5 Derivatives of
Trigonometric Functions
We can find the derivative of tangent x by using the
quotient rule.
d
tan x
dx
cos 2 x  sin 2 x
cos 2 x
d sin x
dx cos x
1
cos 2 x
cos x  cos x  sin x    sin x 
cos 2 x
d
tan  x   sec 2 x
dx
sec 2 x
3.5 Derivatives of
Trigonometric Functions
Derivatives of the remaining trig functions
can be determined the same way.
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
3.5 Derivatives of
Trigonometric Functions
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
da d 2v d 3s
j (t ) 
 2  3
dt dt
dt
3.5 Derivatives of
Trigonometric Functions
3.6 Chain Rule
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
6
dx
dy
2
du
6  23
dy
dy du


dx
du dx
u  3x  5
du
3
dx
3.6 Chain Rule
Chain Rule:
dy dy du


dx du dx
If f g is the composite of y  f  u  and u  g  x  ,
then:

f
g

  fat u  g x  gat x  f ' ( g ( x))  g ' ( x)
example:
f
g  at x  2
f  x   sin x
g  x   x2  4
f   x   cos x
g  x   2x
g  2  4  4  0
cos  0   2  2
1 4
f   0  g   2
Find:
4
3.6 Chain Rule
f  g  x    sin  x 2  4 
y  sin  x 2  4 
y  sin u
u  x2  4
dy
 cos u
du
du
 2x
dx
dy dy du


dx du dx
dy
 cos u  2 x
dx
dy
 cos  x 2  4   2 x
dx
dy
 cos  22  4   2  2
dx
dy
 cos  0   4
dx
dy
4
dx
3.6 Chain Rule
Here is a faster way to find the derivative:
y  sin  x 2  4 
d 2
y  cos  x  4    x  4 
dx
2
y  cos  x 2  4   2 x
At x  2, y  4
Differentiate the outside function...
…then the inside function
3.6 Chain Rule
d
cos 2  3 x 
dx
d
2 cos  3 x    cos  3 x 
dx
d
2 cos  3x    sin  3x    3x 
dx
2cos  3x   sin  3x   3
6cos  3x  sin  3x 
2
d
cos  3 x  
dx
The chain rule can be used
more than once.
(That’s what makes the
“chain” in the “chain rule”!)
3.6 Chain Rule
Derivative formulas include the chain rule!
d n
n 1 du
u  nu
dx
dx
d
du
sin u  cos u
dx
dx
d
du
cos u   sin u
dx
dx
d
du
2
tan u  sec u
dx
dx
etcetera…
3.6 Chain Rule
Find
dy
dx
y  cos(3x 2  x)
dy
  sin( 3 x 2  x)(6 x  1)
dx
y  sin(cos( x))
dy
 cos(cos x)(  sin x)
dx
dy
 3 cos 2 (4 x3  2 x)(  sin( 4 x3  2 x))(12 x 2  2)
dx
dy
 (36 x 2  6) cos 2 (4 x3  2 x)(  sin( 4 x3  2 x))
dx
y  cos3 (4 x3  2 x)
3.6 Chain Rule
The chain rule enables us to find the slope of
parametrically defined curves:
dy dy dx


dt dx dt
dy
dt  dy
dx
dx
dt
The slope of a parametrized
curve is given by:
dy
dy
 dt
dx
dx
dt
3.6 Chain Rule
Example:
These are the equations for
an ellipse.
x  3cos t
dx
dy
 3sin t
 2 cos t
dt
dt
y  2sin t
dy 2 cos t
2

  cot t
dx 3sin t
3
3.7 Implicit Differentiation
x2  y 2  1
d 2 d 2 d
x 
y  1
dx
dx
dx
dy
2x  2 y
0
dx
dy
2y
 2 x
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 2 x

dx 2 y
dy
x

dx
y
3.7 Implicit Differentiation
This can’t be solved for y.
2 y  x 2  sin y
d
d 2 d
dy
2x
2y 
x  sin y

dx
dx
dx
dx 2  cos y
dy
dy
2  2 x  cos y
This technique is called
dx
dx
implicit differentiation.
dy
dy
2  cos y
 2x
dx
dx
1 Differentiate both sides w.r.t. x.
dy
dy
 2  cos y   2 x
2 Solve for
.
dx
dx
3.7 Implicit Differentiation
Implicit Differentiation Process
1.
2.
3.
4.
Differentiate both sides of the equation with respect to x.
Collect the terms with dy/dx on one side of the equation.
Factor out dy/dx .
Solve for dy/dx .
3.7 Implicit Differentiation
Find the equations of the lines tangent and normal to the
curve
x 2  xy  y 2  7 at (1, 2) .
x 2  xy  y 2  7
Note product rule.
dy
 dy

2x   x  y  2 y
0
dx
 dx

dy
dy
2x  x  y  2 y
0
dx
dx
dy
 2 y  x  y  2x
dx
dy y  2 x

dx 2 y  x
2  2  1
22
4
m


2  2   1 4  1
5
3.7 Implicit Differentiation
Find the equations of the lines tangent and normal to the
curve
4
m
5
x 2  xy  y 2  7 at (1, 2) .
tangent:
normal:
4
y  2   x  1
5
4
4
y2 x
5
5
5
y  2    x  1
4
5
5
y2  x
4
4
4
14
y  x
5
5
5
3
y   x
4
4
3.7 Implicit Differentiation
3.7 Implicit Differentiation
d2y
3
2
Find
if
2
x

3
y
7 .
2
dx
y  2 x  x 2 y
 
y
3
2
2x  3 y  7
y2
6 x  6 y y  0
2
6 y y  6 x 2
6 x 2
y 
6 y
x2
y 
y
2x x 2
y 
 2 y
y y
2x x 2 x 2
y 
 2
y y y
2x x 4
y 
 3
y y
Substitute y
back into the
equation.
3.7 Implicit Differentiation
Rational Powers of Differentiable Functions
Power Rule for Rational Powers of x
If n is any rational number, then
d n
x  nx n1
dx
3.7 Implicit Differentiation
Proof: Let p and q be integers with q > 0.
yx
p
q
yq  x p
qy
q 1
Raise both sides to the q power
Differentiate with respect to x
dy
p 1
 px
dx
Solve for dy/dx
3.7 Implicit Differentiation
dy px p1
 q1
dx qy
Substitute for y
dy
px p1

p / q q 1
dx q( x )
p 1
dy
px
 p p / q
dx qx
dy px p1( p p / q )

dx
q
Remove parenthesis
Subtract exponents
dy p ( p / q )1
 x
dx q
3.8 Derivatives of Inverse
Trigonometric Functions
y
8
Slopes are
reciprocals.
y  x2
Because x and y are
reversed to find the
reciprocal function, the
following pattern always
holds:
6
4
 2, 4 
2
m4
 4, 2 
0
0
2
y
1
m
4
4
6
x
8
The derivative of f 1 ( x)
x
Derivative Formula for Inverses:
1
df
dx

x f (a)
1
df
dx
x a
evaluated at f ( a )
is equal to the reciprocal of
the derivative of f ( x )
evaluated at a .
3.8 Derivatives of Inverse
Trigonometric Functions
We can use implicit
differentiation to find:
d
sin 1 x
dx
1.5
y  sin 1 x
1
y  sin x
0.5
y  sin 1 x
-1.5
-1
-0.5
0
-0.5
sin y  x
dy
cos y
1
dx
d
d
sin y 
x
dx
dx
dy
1

dx cos y
-1
-1.5
0.5
1
1.5
3.8 Derivatives of Inverse
Trigonometric Functions
We can use implicit
differentiation to find:
y  sin 1 x
sin y  x
dy
cos y
1
dx
dy
1

dx cos y
d
sin 1 x
dx
sin 2 y  cos2 y  1
d
d
sin y 
x
dx
dx
cos2 y  1  sin 2 y
dy
1

dx
1  sin 2 y
dy
1

dx
1  x2
cos y   1  sin 2 y
But 

 y

2
2
so cos y is positive.
 cos y  1  sin 2 y
3.8 Derivatives of Inverse
Trigonometric Functions
y  sin 1 x
sin y  x
dy
cos y
1
dx
dy
1

dx cos y
dy
1

1
dx cos(sin x)
1
x
1
dy
1

dx
1  x2
sin x
1  x2
3.8 Derivatives of Inverse
Trigonometric Functions
Find
d
tan 1 x
dx
y  tan 1 x
tan y  x
dy
sec y
1
dx
2
dy
1

dx sec 2 y
dy
1

2
1
dx sec (tan x)
1  x2
x
1
dy
1

dx 1  x 2
tan x
1
3.8 Derivatives of Inverse
Trigonometric Functions
Find
d
sec 1 x
dx
y  sec 1 x
sec y  x
dy
1

dx sec(sec 1 x) tan(sec 1 x)
dy
1

dy
sec y tan y
 1 dx | x | x 2  1
dx
dy
1

dx sec y tan y
x
x2 1
sec 1 x
1
3.8 Derivatives of Inverse
Trigonometric Functions
1
cos x 

2
1
1
 sin x cot x 
d
1
du
1
sin u 
dx
1  u 2 dx
d
1 du
tan 1 u 
dx
1  u 2 dx
d
1
du
1
sec u 
dx
u u 2  1 dx

2
1
1
 tan x csc x 

2
 sec 1 x
d
1 du
1
cos u  
dx
1  u 2 dx
d
1 du
1
cot u  
dx
1  u 2 dx
d
1
du
1
csc u  
dx
u u 2  1 dx
3.8 Derivatives of Inverse
Trigonometric Functions
Your calculator contains all
six inverse trig functions.
However it is occasionally
still useful to know the
following:
1
sec x  cos  
x
1
1
cot x 
1

2
 tan 1 x
1
csc x  sin  
 x
1
1
3.8 Derivatives of Inverse
Trigonometric Functions
dy
Find
dx
1
y  cos (3x )
2
1
y  cot  
 x
1
y  x sec1 x
dy
1
6x

(6 x )  
2
2
dx
(1  (3x )
1  9 x4
dy
1  1
1

 2   2
1  x  x 1
dx
1 2
x
dy
1
x
 (sec 1 x)(1)
dx
| x | x2  1
3.9 Derivatives of Exponential and
Logarithmic Functions
Look at the graph of
ye
x
3
2
The slope at x = 0
appears to be 1.
If we assume this to be
true, then:
1
lim
e
h 0
-3
-2
-1
0
-1
1
x
2
0 h
e
1
h
0
3
definition of derivative
3.9 Derivatives of Exponential
and Logarithmic Functions
Now we attempt to find a general formula for the
x
derivative of y  e using the definition.
d x
e xh  e x
e  lim
h 0
dx
h
 
e x  eh  e x
 lim
h 0
h
 x eh  1 
 lim  e 

h0
h 

h

e
1 
x
 e  lim 

h0
 h 
This is the slope at x = 0, which
we have assumed to be 1.
 e 1
x
e
x
 
d x
x
e e
dx

3.9 Derivatives of Exponential
and Logarithmic Functions
e
x
is its own derivative!
If we incorporate the chain rule:
d u
u du
e e
dx
dx
We can now use this formula to find the derivative of
ax
3.9 Derivatives of Exponential
and Logarithmic Functions
 
d x
a
dx
d ln a x
e
dx
 


d x ln a
e
dx
d
x ln a
e   x ln a 
dx
d x
x
a  a ln a
dx
Incorporating the chain rule:
 
d u
du
u
a  a ln a
dx
dx
3.9 Derivatives of Exponential
and Logarithmic Functions
So far today we have:
d u
u du
e e
dx
dx
 
d
du
u
u
a  a ln a
dx
dx
Now it is relatively easy to find the derivative of ln x .
3.9 Derivatives of Exponential
and Logarithmic Functions
y  ln x
y
e x
 
d y
d
e   x
dx
dx
y dy
e
1
dx
dy 1
 y
dx e
d
1
ln x 
dx
x
d
1 du
ln u 
dx
u dx
3.9 Derivatives of Exponential
and Logarithmic Functions
To find the derivative of a common log function, you
could just use the change of base rule for logs:
d
d ln x
1 d
1 1
log x 

ln x 

dx
dx ln10
ln10 dx
ln10 x
The formula for the derivative of a log of any base
other than e is:
d
1 du
log a u 
dx
u ln a dx
3.9 Derivatives of Exponential
and Logarithmic Functions
 
d u
u du
e e
dx
dx
d
du
u
u
a  a ln a
dx
dx
d
1 du
ln u 
dx
u dx
d
1 du
log a u 
dx
u ln a dx
3.9 Derivatives of Exponential
and Logarithmic Functions
Find y’
ye
y 3
2x
y' 2e
x2
2x
y'  3 ln( 3)(2 x)
1
3
2
y '  3 (3x ) 
x
x
1
4x
y' 
(e )( 4)
4x 2
1  (e )
x2
y  ln x
3
1
y  sin (e )
4x
3.9 Derivatives of Exponential
and Logarithmic 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
1 dy
1
 x   ln x
y dx
 x
dy
 y 1 ln x 
dx
dy
 x x 1 ln x 
dx