Transcript Document 7659834

2.2 Fundamental derivation rules
1.Derivation rules for sum, difference, product and quotient
of functions
2. Derivation rules for composite functions
3. The derivative of an inverse function
4.Derivatives for the fundamental elementary functions
5.Higher order derivatives
1.Derivation rules of rational operations
Th2.2.1 I f t he f unct i ons u =u( x) and v = v( x) are deri vabl e at x，
t hen t hei r sum, di f f er ence , pr oduct and quot i ent ar e
al l der i vabl e at x，and
(v( x)  0)
(1) (u  v)  u  v
proof Suppose f ( x )  u( x )  v ( x ) , then
f ( x  x )  f ( x )
f ( x )  lim
x 0
x
'
[u( x  x )  v( x  x )]  [u( x )  v( x )]
 lim
x 0
x
v( x  x )  v( x )
u( x  x )  u( x )
 lim
 lim
x 0
x 0
x
x
 u ( x)  v ( x)
'
'



(
u
v
)

u
v

u
v
(2)
proof
Suppose f ( x)  u ( x)v( x) ,
then
u( x  x )v( x  x )  u( x )v( x )
f ( x  x )  f ( x )
 lim
f ( x )  lim
x  0
x  0
x
x
v( x  x )  v( x ) 
 u( x  x )  u( x )
 lim 
v( x  x )  u( x )

x  0

x

x


 u ( x)v( x)  u ( x)v( x)
corollary 1) ( C u )  C u  ( C is a constant )
2) ( uvw ) 
u vw  uvw  uvw
(1) (2) are also valid for any finite number of functions.
Exampl e Suppose y  e (sin x  cos x )，f i nd y .
x
'
u  u v  u v
(3)   
2
v
v
u ( x)
Proof Suppose f (x)  v ( x ) , then
u( x  x ) u ( x)

v ( x  x ) v ( x )
f ( x  x )  f ( x )
 lim
x  0
x  0
x
x
v ( x  x )  v ( x )
 u(x +x )  u( x )
v (x )  u(x )

x
x
 lim 
x  0
v ( x  x )v ( x )


f ( x )  lim
u ( x) vu((xx))v(ux()x) v( x)

2
v
u( x  x )v ( x )  u( (x)x )v ( x  x )
x v ( x  x )v ( x )





Example 2.2.2-3
证

sin x  (sin x) cos x  sin x (cos x)

(tan x)  

cos 2 x
 cos x 
2
2
cos
x
2

sin
x

sec
x

2
cos x

 cos x
 1   (sin x)

(csc x)  
 
2
2
 sin x 
sin x
sin x
  csc x cot x
Similarly
(cot x)   csc 2 x , (sec x)  sec x tan x .
2. Derivation rules for composite functions
Th2.2.2(chain rule)Suppose that the function u=  ( x)
is derivable at x,and the function y  f (u ) is derivable
at u=  ( x).Then the composite function y  f ( ( x))
is derivable at x ,and
dy
dy dy du
 f (u ) ( x) or

dx
dx du dx
Example2.2.4 Let y  sin( x2  3), find y
The theorem can be generalized to several intermediate
variables.
For instance,
y
u
d y d y d u dv



d x d u dv d x
 f (u )   (v)   ( x)
v
x
Key: Decompose the function into some simple functions
and find derivatives from outside to inside successively.
Example
(2) y  e
sin2
1
x
• P122 T4(4)
Exampl e Pr ove t he der i vat i ve f or mul a f or
t he hyper bol i c f unct i ons
'
'
( shx) =chx , ( chx) =shx ,
1
'
( thx) = 2 ,
ch x
3. The derivative of an inverse function
Th2.2.3 Suppose that x =f ( y ) is a strictly monotone
continuous function on an interval I;if it is derivable at
y and f ( y )  0 ,then its inverse function y =f 1 ( x)is also
derivable at the corresponding x,and
1
(f )( x) 
f ( y )
1
dy
1
or

dx dx
dy
Example 2.2.8. Find the derivatives of the inverse
trigonometric functions.
 
y  ( , ) ,
Solution
For
then
2 2
so cos y  0
1


(sin y ) cos y
1
1  sin 2 y
'
(arcsin x )
by

arccos x   arcsin x
2
similarly
4.Derivatives for the fundamental elementary functions
 1
(C )  0
(sin x)  cos x
(tan x)  sec 2 x
(sec x)  sec x tan x
(a x )  a x ln a
( x  )   x
(cos x)   sin x
2
(cot x)   csc x
(csc x)   csc x cot x
( e x )  e x
1
(log a x) 
x ln a
1
1
(ln x) 
x
(arcsin x) 
1 x
1
(arctan x) 
1  x2
2
(arccos x)  
1
1  x2
1
(arc cot x)  
1  x2
Example Supposef ( x )  u( x )v ( x ) , wher e u( x ) and v ( x )
ar e al l der i vabl e，and u( x )  0, f i nd f ( x ).
Ex . Suppose y  f ( f ( f ( x))) ,
y
where f ( x)exists, find
Solution:
y  f ( f ( f ( x)) ) f ( f (x) )  f (x)
5.Higher -order derivatives
Definition 2.2.1.
If the derivative of y  f (x) i.e. y  f ( x) is derivable,
then its derivative is called the second order derivative of
f (x)
Denoted by
or
i.e.
2
d y d dy

(
)
2
d x dx
dx
In general, the derivative of order n-1 derivative is
called the derivative of order n, and is denoted by
n
d y
(n)
or y
n
dx
i.e.
n
n 1
d y
d d y

( n 1 )
n
dx
d x dx
Example 1.
Suppose
find
Solution 
y  a1  2a2 x  3a3 x 2    nan x n 1
y  2 1a2  3 2a3 x    n(n  1)an x n  2
By means of mathematical induction , we have
y ( n)  n!an
y  e a x , find y (n ) .
Example2. Suppose
3 ax



Solution: y   ae , y   a e , y  a e ,  ,
ax
2 ax
y (n)  a n e ax
x ( n)
(e )
In particular:
e
x
find
Example3. Suppose
1
1
2 1 2
, y   (1)
,
Solution: y  
, y   
2
3
(1  x)
(1  x)
1 x
,
y
(n )
 (1)
n 1
(n  1)!
(1  x) n
0!=1
Example4. Suppose
find
Solution: y  cos x  sin( x  2 )
y  cos( x  2 )  sin( x  2  2 )
 sin( x  2  2 )
y  cos( x  2  2 )  sin( x  3  2 )
so ,
(sin x)
(n)
)
n

 sin( x  2
Bu the similar means:
(cos x) ( n )  cos( x  n  2 )
The operation rules of higher-order derivatives(Th2.2.4)
suppose
and
are both derivable of order n , th
n(n  1)
2!
n(n  1)  (n  k  1)

k!
Leibniz formula
Example 2.2.11 Let f ( x)  x e , find f
2 2x
(20)
 x
P123T11 Let f ( x)  3x 3  x 2 x , find the highest order n such
that f ( n ) (0) exists.
3

4
x
,
f
(x
)


Solution:
2 x 3 ,
x0
x0
2 x3  0
 f  (0)  lim
2
0
x

0
12
x
,

x
x 0



f
(x
)

 2
4 x3  0
6x , x  0


f  (0)  lim
0

x
x 0
6x2
while f (0)  lim
24x , x  0
0

 x
x 0
 f (x)  
 12x , x  0
12 x 2
0
f (0)  lim
x
x 0
but f (0)  12 , f (0)  24 ,  f (0) does not exists.
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The method of finding higher order derivatives:
(1) Find the derivative order by order
(2) By means of mathematical induction
(3) By Leibniz formula
(4) From the known derivative formula
For instance,
n!
1 (n)
n

  (1)
ax
(a  x) n 1
n!
1 (n)

 
ax
(a  x) n 1
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Example P123T10(4) y 
1
x 2  3x  2
1
1


x  2 x 1
y
(n)


1
1
 (1) n ! 

n 1
n 1 
( x  1) 
 ( x  2)
n
1
Example2.2.12 Let f ( x) 
, find f ( n )  x 
x( x  1)
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Exercise in class
1. How to find the derivative of order n for the following
functions?
3
(1)
x
y
1 x
y
(n)
n!

, n3
n 1
(1  x)
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( 2) y  cos 2 wx
1  cos 2 wx ( n )
y (
)
2
1

n

  2  cos  2 x  n 
2
2

(n)
2.3 Derivation of implicit functions and functions
defined by parametric equations
1.Derivation of implicit functions
2.Method of derivation of a function defined by
parametric equations
3.Related rates of change
1.Derivation of implicit functions
explicit function.
If the function
is defined by the equation
then it is called an implicit function .
For instance,
can define a function with the dependent variable y and
independent variable x ,but it can not be explicited.
How to find the derivative of an implicit function?
Taking derivative on both sides with respect
to x
the equation about
y
Example2.3.1 Find the derivative of the function y=f(x)
defined by the equation y=1+xsiny.
Example2.3.2 Suppose that the function y  y ( x )
is defined by the equation e y  xy  e;
dy
find
dx
2
d y
, 2
x  0 dx
x 0
Example. Find the tangent line of
at
Solution: Taking derivatives on both sides respect to x
x 2
 y  y  0
8 9
3
9 x
 y x  2  
x2  
4
16 y y  3 3
3
y2 3
2
3
3
( x  2)
So the tangent line is y  3  
2
4
i.e.
The method of logarithmic derivation.(P133T3)
Example Find the derivative of
Solu: Taking the natural logarithm on both sides
Taking derivative on both sides respect to x

1
sin
x

y  cos x  ln x 
y
x
sin x
sin x
y   x (cos x  ln x 
)
x
Note:
1) For a powerexponent function y  u v ,(u  0)
ln y  v ln u
1
u v
y   v ln u 
y
u
u v
v
y   u ( v ln u 
)
u
2) Some explicit functions .
a
ln y  x ln  a [ ln b  ln x ] b [ ln x  ln a ]
b
a a b
y
 ln  
b x x
y
( x  1)( x  2)
and, y 
( x  3)( x  4)
u
( ln u ) 
u
1
ln y   ln x  1  ln x  2  ln x  3  ln x  4
2
y 1 1
1
1
1
 



y 2 x 1 x  2 x  3 x  4



1
1
1
1



x 1 x  2 x  3 x  4

2.Method of derivation of a function defined by
parametric equations
If the parametric equation
y=f(x),
can define a function
is derivable and
then
If  (t )  0 we have
d y d y d t  d y  1   (t )


dx d t dx d t dx  (t )
dt
If
 (t )  0
dx dx d t dx 1
 (t )
 
 

d y d t d y d t d y  (t )
dt
(here means x is the function of y )
if
are both twice derivable, and
then
is twice derivable.
From the new parametric equation
x   (t )
d y  (t )

dx  (t )
d d y dx
d 2 y d (d y )
 ( )

2
dx dx
d t dx d t
dx
 (t ) (t )   (t ) (t )

  (t )
2
 (t )
 (t ) (t )   (t ) (t ) yx  xy


3
x 3
  (t )
,we have
Note : If
?
Example 2.3.4Find the derivatives of function
defined by the cycloid
 x  a ( t  sin t )

 y  a (1  cos t )
2

x

t
 2t
Example. Suppose that 
(0    1)
2
 t  y   sin y  1
can define the function y  y (x) , find
Solution: Taking derivatives on both sides
dx
dx
 2 (t  1)
 2t  2
dt
dt
dy
2t
dy
dy

  cos y
2t 
0
d t 1   cos y
dt
dt
so
dy
t
dy
 d t dx 
(t  1)(1   cos y )
dx
dt
3.related rates of change
are two derivable functions
are dependent
are dependent too
is called the related rates of change.
Solving the problem of related rates of change:
Find the equation F(x,y)=0
Taking derivative on both sides respect to t
Get the expression between x(t ) and y(t )
Find the rate that we want
Example2.3.5 Water run s in to a vertical cylindrical tank
with radius 5cm from a conical tank of altitude 18cm and
base radius 6cm.Suppose that initially the conical tan k is
full of water.How fast is the water level rising in the
cylindrical tan k when the water is 12cm deep and the water
level in t he conical tan k is falling at the rate of 1cm / s ?
Example2.3.6 A hot  air ballon rises up vertically from the earth
and is tracked by arange finder 500m from the lift off po int.
At the moment when the angle of elevation of the range finder

is , the angle is in crea sin g at the rate of 0.14rad / min.
4
How fast is the balloon ri sin g at that moment ?
Have a think :Suppose
Solution 1.
1 e
find the derivative of
the inverse function .
x
Solution2
Taking derivatives on both sides with respcet to y