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Advanced Mathematics D
Chapter Three
The Derivatives
Definition of The Derivative Function
 The function f’ defined by the formula
f ( x  h)  f ( x )
f '( x)  lim
h 0
h
which is called the derivative of f with respect to x.
The domain of f’ consists of all x in the domain
of f for which the limit exists.
Definition of Differentiability
 A function f is said to be differentiable at x0 if
the limit
f ( x0  h)  f ( x0 )
f '( x0 )  lim
h 0
h
exists. If f is differentiable at each point of the
open interval of the form (a,b), then we say
that it is differentiable on (a,b), and similarly for
open intervals of the form (a,+∞) and (-∞,+∞).
The last case, we say that f is differentiable
everywhere.
Differentiable Point
VS Continuous Point
Theorem
If a function f is differentiable at x0, then f is
continuous at x0
One-sided Derivative
 Left-hand derivatives
f ( x0  h)  f ( x0 )
f  '( x0 )  lim
h 0
h
 Right-hand derivatives
f ( x0  h)  f ( x0 )
f  '( x0 )  lim
h 0
h
Differentiable in a
Cloed/Half Closed Interval
F is differentiable on an interval of the form
[a,b], [a,+∞),(-∞,b],[a,b) of (a,b] if it is
differentiable all points inside the interval
and the appropriate one-sided derivative
exists at each included endpoint.
Other Notations

d
f '( x)  [ f ( x)],
dx
f '( x)  Dx [ f ( x)],
 For y = f (x), f '( x)  y '( x),
dy
f '( x) 
dx
 Derivative at point x0,
d
f '( x0 )  [ f ( x)]
,
dx
x  x0
f '( x0 )  y '( x0 ),
f '( x0 )  Dx [ f ( x)] x  x
dy
f '( x0 ) 
dx
0
x  x0
Derivative of a Constant
 f (x) = const. ═> f’(x) = 0
 Theorem
The derivative of a constant function is 0;
that is, if c is any real number, then
d
[c ]  0
dx
Derivatives of Integer Powers of x
 Theorem (The Power Rule)
If n is any integer, then
d n
[ x ]  nx n 1
dx
Derivative of a Constant Times a
Function
 Theorem (Constant Multiple Rule)
If f is differentiable at x and c is any real
number, then cf is also differentiable at x and
d
d
[cf ( x)]  c [ f ( x)]
dx
dx
Derivatives of Sums and Differences
 Theorem (Sum and Difference Rules)
If f and g are differentiable at x, so are f±g
d
d
d
[ f ( x)  g ( x)]  [ f ( x)]  [ g ( x)]
dx
dx
dx
Derivative to Polynomia
Joint together about rules, we have
d
d
n
n 1
1
[ Pn ( x)]  [an x  an 1 x  ...  a1 x  a0 ]
dx
dx
n 1
n2
 an nx  an 1 (n  1) x  ...  a1
High Derivatives
 f’ is a function
 If f’ is derivatiable, the derivative of f’ is denoted
f’’, called the second derivative of f
 So on, y '  dy  d [ f ( x)]
dx dx
2
d2y d  d
 d
y ''  2 
f ( x)   2 [ f ( x)]

dx
dx  dx
 dx
......
y(n)
dny dn
 n  n [ f ( x)]
dx
dx
Derivative of A Product
At first view
If (fg)’=f’g’?
The answer is negative
Theorem (The Product Rule)
 If f and g are differentiable at x, then so is the
product f*g and
d
d
d
[ f ( x) g ( x)]  f ( x) [ g ( x)]  g ( x) [ f ( x)]
dx
dx
dx
Derivative of a Quotient
 If f and g is differentiable at x and g(x)≠0, then
f/g is differentiable at x and
d
d
g ( x) [ f ( x)]  f ( x) [ g ( x)]


d f ( x)
dx
dx

2
dx  g ( x) 
 g ( x) 
Derivatives of Trigonometric Function

d
d
sin x   cos x,
cos x    sin x
dx
dx
d
d
2
 tan x   sec x,
sec x   sec x tan x
dx
dx
d
d
2
cot x    csc x,
csc x    csc x cot x
dx
dx
The Derivatives of Compositions
 An Example: how to find
d
( x 2  1)100  ?
dx
 We can consider
y  u , & u  x 1
100
2
Chain Rule
 Theorem
 If g is differentiable at x and f is differentiable
at g(x), then the composition f◦g is
differentiable at x. Moreover, if
y  f ( g ( x)) & u  g ( x)
 y  f (u ) and
dy dy du

*
dx du dx
Other Expression of the Chain Role

d
 f ( g ( x))  ( f g ) '( x)  f '( g ( x)) g '( x)
dx
d
du
 f (u )  f '(u )
dx
dx
Generalized Derivative Formulas
1 du
d  
d n
n 1 du
u   nu

u
,
dx   2 u dx
dx
dx
du
d
du
d
cos u    sin u
sin u   cos u ,
dx
dx
dx
dx
du
d
du
d
2
sec u   sec u tan u
 tan u   sec u ,
dx
dx
dx
dx
du
d
du
d
2
csc u    csc u cot u
cot u    csc u ,
dx
dx
dx
dx
Related Rates
It is in fact a derivative of component
function
Known a function’s derivative with respect
to time t (change rate)
Find the change rate of another function
which is a component function of previous
function
Strategy for Solving Related Rate
Problem
1. Assign letters . Give a definition for each
letters
2. Identity the rates of change. Interpret each
rate as a derivative
3. Find an equation
4. Differentiate both sides of the equation with
respect to t
5. Substitute all known variables
Local Linear Approximation
 Value of f (x) near x0 can be approximated by
f(x0) plus the distance multiply the derivative of
f(x) at x0, i.e.
f ( x)  f ( x0 )  f '( x0 )( x  x0 )
Error in Local Linear Approximations
Linear approximations of x to x0 has an
error
The error depends on
The distance to x0
The high derivatives of f (x0)
Differentials
difference:
y  f ( x0 )  f ( x)  f '( x)( x0  x)  o(| x0  x |)
 f '( x)x  o(x)
limit: differentials:
derivative:
dy  f '( x) dx
dy
 f '( x)
dx
Derivative vs Differential
d
[c ]  0
dx
d
df
[cf ]  c
dx
dx
d
df dg
[ f  g] 

dx
dx dx
d
dg
df
[ fg ]  f
g
dx
dx
dx
d  f  1  df
dg 
 2 g
f



dx  g  g  dx
dx 
d [c ]  0
d [cf ]  cdf
d [ f  g ]  df  dg
d [ fg ]  fdg  gdf
f 1
d    2  gdf  fdg 
g g
Implicit Function
Definition
We will say that a given equation in x and y
defines the function f implicitly if the graph
of y=f (x) coincides with a portion of the
graph of the equation.
Derivatives of Rational Powers of x
 It has the same form as the power is integer
d r
r 1
[ x ]  rx
dx
For r is a rational number
Derivatives of Logarithmic Function

d
1
[ln x] 
dx
x
d
1
[log b x] 
dx
x ln b
d
1 du
[ln u ] 
dx
u dx
Differentiability for Inverse function
Theorem
Suppose that the domain of a function f is
an open interval I on which f’(x) >0 or on
which f’(x) <0. Then f is one-to-one, f -1(x)
is differentiable at all values of x in the
range of
d 1
1
[ f ( x)] 
dx
f '( f 1 ( x))
Differentiability for Inverse function
Theorem
Suppose that the domain of a function f is
an open interval I and that f is differentiable
and one-to-one in this interval, then f -1(x) is
differentiable at any point x in the range of f
at which f (f -1(x))≠0, and
d 1
1
[ f ( x)] 
dx
f '( f 1 ( x))
Derivatives of Exponential Function

d x
x
[e ]  e
dx
d x
x
[b ]  b ln b
dx
d u
u du
[e ]  e
dx
dx
Derivatives of Trigonometric Function

d
1
d
1
1
1
[sin x] 
,
[cos x] 
dx
dx
1  x2
1  x2
d
1
d
1
1
1
[tan x] 
,
[cot x] 
2
dx
1 x
dx
1  x2
d
1
d
1
1
1
[sec x] 
,
[csc x] 
2
dx
x x  1 dx
x x2 1
L’HÔPITAL’s Rule
Theorem of 0/0
Suppose that f and g are differentiable functions
on an open interval containing x=a. except
possible at x=a and that
f '( x)
lim f ( x)  lim g ( x)  0, lim
exists or =  
x a
x a
x a g '( x)
Then
f ( x)
f '( x)
lim
= lim
x a g ( x)
x a g '( x)

x

a
, x  .
The statement is also true for
Applying L’HÔPITAL’s Rule
Step 1 Check that the limit f(x)/g(x) is an
indeterminate form of type 0/0
Step 2 Differentiate f and g separately
Step 3 Find that limit of f’(x)/g’(x)
If the limit is finite or±∞, it is equal to limit
of f(x)/g(x)
L’HÔPITAL’s Rule
Theorem of ∞/∞
Suppose that f and g are differentiable functions
on an open interval containing x=a. except
possible at x=a and that
f '( x)
lim f ( x)  lim g ( x)  , lim
exists or =  
x a
x a
x a g '( x)
Then
f ( x)
f '( x)
lim
= lim
x a g ( x)
x a g '( x)

x

a
, x  .
The statement is also true for
Indeterminate Forms
0/0, ∞/ ∞, 0∙ ∞, ∞± ∞, 1∞
Method to determine the limit:
 Step 1 Transfer the form into 0/0 or ∞/∞
 Step 2 Using L’HÔPITAL’s Rule