§
2.1( 2.7) Derivatives and Rates of Change 導數和變化率
−
Def
The tangent line (切線）to the curve y = ƒ(x) at the point P(a, ƒ(a)) is the line through P with slope（斜率）
if the limit exists.
a x
a f x
mPQ f
−
= −
) ( )
( (the slope of the secant line)
h
a f h
a
mPQ f ( + ) − ( )
=
h
a f h
a f a
x
a f x
m f
m PQ x a h
P Q
) ( )
lim ( )
( )
lim (
lim 0
−
= +
−
= −
= → → →
[Ex1][Ex1]
Find an equation of the tangent line to the parabola at the point P (1,1) [Sol 1]:
Here we have a = 1 and ƒ(x) = , so the slope of the tangent line isx2 x2
y =
2
1 1 1
( ) (1) 1
lim lim lim( 1) 2
1 1
x x x
f x f x
m x
x x
→ → →
− −
= = = + =
− −
Using the point-slope formula for a line , we have that the
equation of the line with slop 2 passing through the point (1,1) is
（方程式）
[Sol 2]:
Alternatively, we may use the other formula for the slope:
2 2
0 0 0 0
(1 ) (1) (1 ) 1 1 2 1
lim lim lim lim(2 ) 2
h h h h
f h f h h h
m h
h h h
→ → → →
+ − + − + + −
= = = = + =
Using the point-slope formula , we have that an equation of the tangent line is 1 2( 1) or 2 1
y− = x− y= x−
1 2( 1) or 2 1 y− = x− y= x−
[Ex2][Ex2]
Find an equation of the tangent line to the hyperbola at the point (3,1) [Sol 1 ]:
Here we have a = 3 and ƒ(x) = , then the slope of the tangent is3 x
y 3
= x
3 3 3 3
3 3
( ) (3) 1 1 1
lim lim lim lim ( )
3 3 3 3
x x x x
x
f x f x x
m → x → x → x → x
− −
= − = = = − = −
− − −
Therefore , an equation of the line with slop passing through (3,1) is
[Sol 2 ]:
Using the formula to solve the problem.
As an exercise , you should do it on your own.
1
−3
0
(3 ) (3) limh
f h f
m → h
+ −
= 1 1 ( 3)
y− = − 3 x−
Velocities (速度)
Suppose an object moves along a straight line（直線）according to an equation of motion which is called the position function（位置函數）of the object.
In the time interval from t = a to t = a + h , the )
(t s = f
average velocity ( ) ( )
PQ
displacement f a h f a
time h m
+ −
= = =
We define the velocity (or instantaneous velocity瞬時速度) v(a) at time a to be
h
a f h
a f
h
) ( )
lim (
0
−
= + ( ) →
v a
[Ex3][Ex3]
Suppose that a ball is dropped from the upper observation desk of the CN tower , 450 m above the ground.
(a) What is the velocity of the ball after 5 seconds ?
(b) How fast is the ball traveling when it hits the ground ? [Sol ]:
(the height measured from the ground) ( ) 450 4.9 2
f t = − t
2 2
( ) ( ) [450 4.9( ) ] [450 ]
( ) lim f a h f a lim a h a lim 4.9(2 ) 9.8
v a + − − + − − a h a
= = = ⋅⋅⋅ = − + = −
0 0 0
( ) lim lim lim 4.9(2 ) 9.8
h h h
v a a h a
h h
→ → →
= = = ⋅⋅⋅ = − + = −
(a) The velocity after 5s is
(the minus sign means that it is going downward.) (5) 9.8 5 49(m )
v
= − ⋅ = − s
(b) When the ball hit the ground , ƒ( ) = 0.
That is ,
So the velocity as it hits the ground is
2
0 0
450 4.9 0 450 9.6 s
4.9
t t
− = ⇒ = ≈
t0
0 0
( ) 9.8 9.8 450 94 4.9
v t t m
= − = − ≈ s
Other Rates of Change 其他變化率
Suppose y = ƒ(x) [the quantity y depends on another quantity x]
As x changes from to , the change in x :
the change in y :
The average rate of change of y w.r.t. x
x1
x2
1
2 x
x x = −
∆
) ( )
(x2 f x1 f
y = −
∆
( ) ( ) f x f x
y −
∆
（平均變化率）
2 1
2 1
( ) ( ) f x f x y
x x x
∆ −
= =
∆ −
We define the instantaneous rate of change of y w.r.t. x at x = to be
which is also the slope of the tangent line to the curve y = ƒ(x) at x1
1 1
( , ( ))x f x
( )
2 1 2 1
2 1
2 1
lim lim ( )
x x x x
f x f x
y
x x x
→ →
∆ −
∆ = −
（瞬時變化率） （對於）
Def
(1) The derivative of a function ƒ at a , denoted by , is
if the limit exists.
(2) f is called differentiable at x=a, if exists.
( ) f a′
0
( ) ( ) ( ) ( )
( ) lim or lim
h x a
f a h f a f x f a
f a → h → x a
+ − −
′ =
−
0
( ) ( )
( ) lim
h
f a h f a
f a → h
+ −
′ =
Notice:
（可微分的）
Notice:
The derivative , , has several meanings :
(1) The slope of the tangent line to the curve y = ƒ(x) at the point (2) The instantaneous rate of change of y = ƒ(x) w.r.t. x when x = a
(3) If ƒ(t) is the position function of a particle that moves along a straight lines , then is the velocity of the particle at time t = a
( ) f a′
( , ( ))a f a
（瞬時變化率）
( ) f a′
[Ex4][Ex4]
Find the derivative of the function at a.
[Sol]:
( ) 2 8 9 f x = x − x+
2 2
( ) ( ) ( 8 9) ( 8 9)
'( ) lim lim
( )( 8)
lim lim( 8) 2 8
x a x a
x a x a
f x f a x x a a
f a x a x a
x a x a
x a a
x a
→ →
→ →
− − + − − +
= = = ⋅ ⋅ ⋅
− −
− + −
= = + − = −
− [Ex5][Ex5]
[Ex5][Ex5]
Find an equation of the tangent line to the parabola at (3,-6) [Sol]:
Let , then (by Ex 4) The slope of the tangent line is
Hence , an equation of the tangent line is
9
2 8
+
−
= x x
y
( ) 2 8 9
f x = x − x + f a′( ) = 2a − 8
(3) 2 3 8 2
f ′ = ⋅ − = −
(-6) 2( 3) or 2
y − = − x − y = − x
[Ex]
The position of a particle is given by the equation of motion s = ƒ(t) = , where t is measured in seconds and s in meters. Find the velocity after 2 seconds.
[Sol]:
+ t 1
1
2 2 2 2
3 (1 )
1 1
( ) (2) 1 3 3(1 ) 2
'(2) lim lim lim lim
2 2 2 ( 2) 3(1 )
1 1
t t t t
t
f t f t t t
f → t → t → t → t t
− +
− + − + −
= = = =
− − − − ⋅ +
−
Thus , the velocity after 2 seconds is 1
(2) ( )
9 f m
′ = − s
2
1 1
lim
3(1 ) 9
t→ t
= − = −
+
§
2.2 (2.8) The Derivative as a Function 導函數
0
( ) ( )
( ) lim
h
f x h f x
f x → h
+ −
′ = The derivative of ƒ :
{ | ( ) exists }
f f
D ′ = x f ′ x ⊆ D [Ex1][Ex1]
Use the graph of ƒ to sketch the graph of ƒ’
[Ex2][Ex2] (a) If , find a formula for ƒ’(x) . (b) Illustrate by comparing the graphs of ƒ and ƒ’
[Sol]:
(a) (b)
0
3 3
0
( ) ( )
( ) lim
[( ) ( )] [ ]
lim
h
h
f x h f x f x
h
x h x h x x
h
→
→
+ −
′ =
+ − + − −
=
x x
x
f = −
) 3
(
0
h→ h
3 2 2 3 3
0
2 2 3
0
2 2 2
0
[ 3 3 ]
lim
3 3
lim
lim(3 3 1) 3 1
h
h
h
x x h xh h x h x x
h
x h xh h h
h
x xh h x
→
→
→
+ + + − − − +
=
+ + −
=
= + + − = −
[Ex] If , find the derivative of ƒ and state the domain of ƒ’.
[Sol]:
( ) 1
f x = x−
0
0
0
( ) ( )
( ) lim
1 1
lim
( 1 1)( 1 1)
lim
( 1 1)
h
h
h
f x h f x
f x h
x h x
h
x h x x h x
h x h x
→
→
→
+ −
′ =
+ − − −
=
+ − − − + − + −
= + − + −
0
0
0
( 1) ( 1)
lim ( 1 1)
lim ( 1 1)
lim 1
1 1
1
2 1
h
h
h
x h x
h x h x
h
h x h x
x h x
x
→
→
→
+ − − −
=
+ − + −
=
+ − + −
=
+ − + −
=
−
{ | 1} [1, ) { | 1} (1, )
f f
D x x
D x x
′
= ≥ = ∞
= > = ∞
[Ex4][Ex4] Find ƒ’ if [Sol]:
( ) 1 2 f x x
x
= − +
0
0
( ) ( )
( ) lim
1 ( ) 1
2 ( ) 2
lim
h
h
f x h f x
f x h
x h x
x h x
h
→
→
+ −
′ =
− + −
+ + − +
=
0
(1 )(2 ) (1 )(2 )
limh (2 )(2 )
x h x x x h
h x h x
→
− − + − − + +
= 0 + + +
2 2
0
0
0
2
(2 )(2 )
(2 2 2 )(2 2 )
lim (2 )(2 )
lim 3
(2 )(2 )
lim 3
(2 )(2 )
3 (2 )
h
h
h
h
h x h x
x h x x xh x x x h xh
h x h x
h
h x h x
x h x
x
→
→
→
→
+ + +
− − + − − − + − + −
= + + +
= −
+ + +
= −
+ + +
= −
+
Other Notations
) ( )
( )
( '
) (
' f x Df x D f x
dx d dx
y df dx
d dx
y dy x
f = = = = = = = x
, x, : D D d
dx differentiation operators 微分算子
) ( )
( '
) (
' f a
dx x d
dx f d dx
df dx
y dy a
f
a x a
x a
x a
x = = = ≠
=
=
=
=
=
) ( )
(x D f a
f
Dx x a ≠ x
= =
Def
A function ƒ is differentiable(可微分的) at a if exists
It is differentiable on an open interval (a,b) if it is differentiable at every point in the interval
Note:
If a function ƒ is differentiable at x = a , then the slope of the tangent line to the curve y = ƒ(x) at (a, ƒ(a)) exists. Therefore , the curve y = ƒ(x) has a tangent line at x = a
( ) ( ) lim
x a
f x f a x a
→
−
−
[Ex5][Ex5] Where is the function ƒ(x) = |x| differentiable （可微分）? [Sol]:
( i ) If x > 0 ,
(ii) If x < 0 ,
0 0
( ) ( )
( ) lim lim
h h
x h x
f x h f h
f x → h → h
+ −
+ −
′ = =
0 0 0
( )
'( ) lim lim lim 1
h h h
x h x x h x h
f x → h → h → h
+ − + −
= = = =
0 0 0
( ) ( )
'( ) lim lim lim 1
h h h
x h x x h x h
f x → h → h → h
+ − − + − − −
= = = = −
is differentiable on (0, ).f
∴ ∞
is differentiable on (f ,0).
∴ − ∞
(iii) At x = 0,
is differentiable on (f ,0).
∴ − ∞
0 0
(0 ) (0)
(0) lim lim
h h
f h f h
f → h → h
+ −
′ = =
0 0 0 0
lim lim 1 lim lim 1
h h h h
h h h h
h h h h
+ + − −
→ → → →
= = ≠ = − = −
∵
Therefore , ƒ(x) is not differentiable at x = 0
0 0
(0 ) (0) (0 ) (0)
. . lim lim
h h
f h f f h f
i e → + h → − h
+ − + −
≠
0
(0 ) (0)
(0) lim D.N.E
h
f h f
f → h
+ −
′ =
∴
The graph has a corner at x = 0, resulting in i.e. ƒ’(0) does not exist
i.e. the slope of the tangent line at (0,0) does not exist.
In this case , the tangent line at (0,0) does not exist.
0 0
(0 ) (0) (0 ) (0)
lim lim
h h
f h f f h f
h h
+ −
→ →
+ − + −
≠
[Ex] Is differentiable at x = 2?
[Sol]:
( ) 2 4 f x = x −
2
0 2 2
4 0
(2 ) (2) ( ) (2)
'(2) lim lim lim
2 2
h x x
f h f f x f x
f → h − x − x
− −
+ − −
= = =
− −
2 2
2 2 2
4 4
lim lim lim ( 2) 4
2 2
x x x
x x
x
x x
+ + +
→ → →
− −
= = + =
− −
2 4 ( 2 4)
lim lim lim ( 2) 4
x x
x
− − −
= = − + = −
2 2 2
4 ( 4)
lim lim lim ( 2) 4
2 2
x x x
x x
x
x x
− − +
→ → →
− − −
= = − + = −
− −
i.e. ƒ’(2) does not exist
i.e. ƒ(x) is not differentiable at x = 2
2 2
( ) (2) ( ) (2)
lim lim
2 2
x x
f x f f x f
x x
+ −
→ →
− −
− ≠ −
∴
The graph has a corner at x = 2 , resulting in
i.e. ƒ’(2) does not exist
⇒ ƒ(x) is not differentiable at x = 2 and
the slope of the tangent line does not exist.
2 ) ( )
lim ( 2
) 2 ( )
lim (
2
2 −
≠ −
−
−
−
+ →
→ x
x f x
f x
f x
f
x x
the slope of the tangent line does not exist.
In this case , the tangent line does not exist.
Corner(折點) => not differentiable
Differentiable functions always have smooth graphs.
Theorem
If ƒ is differentiable at x = a , then ƒ is continuous at x = a
[Proof]:
The given information is that ƒ is diff. i.e. exists.
To prove that ƒ is continuous at x = a , we have to show that lim ( ) ( )
x a
f x f a
→ =
( ) ( ) ( ) lim
x a
f x f a
f a → x a
′ = −
− ( ) ( )
lim ( ) lim[ f x f a ( ) ( )]
f x − x a f a
= ⋅ − +
−
lim ( ) lim[ ( ) ( )]
x a x a
f x x a f a
x a
→ = → ⋅ − +
−
( ) ( ) ( ) ( )
lim lim( ) lim ( ) ( lim ,lim( ),lim ( ) exist) ( ) ( )
lim lim( ) lim ( )
'( ) 0 ( ) ( )
x a x a x a x a x a x a
x a x a x a
f x f a f x f a
x a f a x a f a
x a x a
f x f a
x a f a
x a
f a f a
f a
→ → → → → →
→ → →
− −
= ⋅ − + −
− −
= − ⋅ − +
−
= ⋅ +
=
∵
Therefore , ƒ(x) is continuous at x = a.
Being differentiable => being continuous
being continuous => being differentiable
That is , continuous functions are not necessarily differentiable.
For instance , ƒ(x) =
ƒ(x) is continuous at x = 2 But its graph has a corner at x = 2,
2 4
x −
2
2 2
lim ( ) lim 4 0 (2)
x x
f x x f
→ = → − = = ∴
∵
( ) (2)
'(2) lim f x f does not exist.
f −
= but
But its graph has a corner at x = 2,
Therefore ƒ(x) is not differentiable at x = 2 (its graph is not smooth at x = 2) '(2) lim2 does not exist.
2
x
f = → x
−
Although ƒ is continuous at x=a , it fails to be diff. if its graph is not smooth at x=a Since ,
we have
Therefore , if ƒ is not continuous at x = a , it cannot be differentiable at x = a being not continuous => being not differentiable
being differentiable => being continuous
[Ex]
(1) Find the discontinuities of ƒ.
(2) Find the points at which ƒ is not differentiable (3) Graph the function (find asymptotes first).
[Sol]:
(1) From the definition of ƒ , the suspect discontinuities are x = 0 , x = 4 , x = 5.
0, 0
( ) 5 , 0 4
1 , 4
5
x
f x x x
x x
≤
= − < <
≥
−
(i) At x = 0 , ∴ ƒ is not continuous at x=0
(ii) At x = 4 ,
∴ ƒ(x) is continuous at x = 4.
(iii) ƒ is not continuous at x = 5 since ƒ(5) is undefined.
So the discontinuities of ƒ are x = 0 , x = 5.
0 0 0
lim ( ) 0 lim ( ) lim (5 ) 5
x x x
f x f x x
− + +
→ = ≠ → = → − =
∵
4 4 4 4
lim ( ) lim (5 ) 1 lim ( ) lim 1 1 5
x x x x
f x x f x
x
− − + +
→ → → →
= − = = = =
− lim ( )4 1 (4)
x
f x f
→ = =
∵
4
( ) (4)
limx 4
f x f x
→
−
−
4 4 4
( ) (4) (5 ) 1 4
lim lim lim 1
4 4 4
x x x
f x f x x
x x x
− − +
→ → →
− − − −
= = = −
− − −
∵
(2) Again the suspect points at which f fail to be differentiable are x=0 , x=4 , x=5 (i) At x = 0 , x = 5, ƒ is not continuous. So ƒ is not differentiable at x = 0 ,x = 5.
(ii) At x = 4, let’s compute ƒ’(4) = to find the answer.
1 1 (5 )
( ) (4) 5 1 5 1
x
f x f x x
− −
− − − −
i.e. ƒ’(4) does not exist.
Therefore ƒ(x) is not differentiable at x = 4
4 4 4 4
( ) (4) 5 1 5 1
lim lim lim lim 1
4 4 4 5
x x x x
f x f x x
x x x x
+ + + +
→ → → →
− − − −
= = = =
− − − −
4 4
( ) (4) ( ) (4)
lim lim
4 4
x x
f x f f x f
x x
+ −
→ →
− −
− ≠ −
∴
(3) (i) Vertical asymptotes : ( the potential vertical asymptote is x = 5 ) and
∴ x = 5 is the vertical asymptote of y = ƒ(x) (ii) horizontal asymptotes :
∴ y = 0 is the horizontal asymptote of y =ƒ(x) (iii)
5 5
lim ( ) lim 1 5
x x
f x x
+ +
→ = → = − ∞
∵ −
5 5
lim ( ) lim 1 5
x x
f x x
− −
→ →
= = ∞
−
lim ( ) lim 1 0
5
x x
f x x
→ ∞ = → ∞ =
−
∵
Discontinuity : x = 0 , x = 5
The points at which f fails to be differentiable :
x = 0 (discontinuity) x = 4 (corner)
x = 5 (discontinuity)
[Ex]
Prove that is not differentiable at x = 0 and explain geometrically.
[Poof]:
To prove that ƒ(x) is not diff. at x = 0 , all we need to do is to show that ƒ’(0) does not exist.
( ) 1 3
f x = x
0
( ) (0) '(0) lim
0
x
f x f
f → x
= −
−
1 3 0
0 2 3
lim 0 lim 1
x
x
x x x
→
→
= −
=
= ∞
The limit does not exist (ƒ’(0) does not exist) which means ƒ(x) is not differentiable at x = 0.
The curve is smooth at (0,0), so the tangent line exists.
However, its tangent line at (0,0) is a vertical line whose slope is undefined.
(or the slope does not exist).
This indicates that ƒ’(0) does not exist.
Therefore , ƒ(x) is not differentiable at x = 0.
Therefore , ƒ(x) is not differentiable at x = 0.
Three possibilities for ƒ to fail to be differentiable at x = a :
A discontinuity A corner A vertical tangent ( not smooth ) ( not smooth ) ( smooth ) ( not smooth ) ( not smooth ) ( smooth )
( ) ( ) ( ) ( )
lim lim
( ) does not exist.
The tangent line does not exist.
x a x a
f x f a f x f a
x a x a
f a
+ −
→ →
− −
− ≠ −
′
∴
∵
( ) ( ) ( ) lim
The slope of the tangent line
The tangen
does not exist.
t line exists.
x a
f x f a
f a → x a
′ = − = ±∞
−
Exercises :
1. Find the derivative of ƒ(x) = by using the definition of a derivative.
2. Is ƒ(x) = |x-1| diff. at x = 1? Prove your answer and give geometrical explanation.
− x 4
0 0
0
0 0
0
4 ( ) 4
( ) ( )
(1) ( ) lim lim
( 4 4 )( 4 4 )
lim
( 4 4 )
(4 ) (4 )
lim lim
( 4 4 ) ( 4 4 )
1 1
lim
( 4 4 )
h h
h
h h
h
x h x
f x h f x f x
h h
x h x x h x
h x h x
x h x h
h x h x h x h x
x h x
→ →
→
→ →
→
− + − −
+ −
′ = =
− − − − − − + −
= − − + −
− − − − −
= =
− − + − − − + −
− −
= =
− − + − 2 4 − x
[Sol]:
1 1
1 1
1 1 1
( ) (1) ( 1) 0
(2) lim lim 1
1 1
( ) (1) ( 1) 0
lim lim 1
1 1
( ) (1) ( ) (1) ( ) (1)
lim lim lim D.N.E
1 1 1
. . (1) D.N.E is not d
x x
x x
x x x
f x f x
x x
f x f x
x x
f x f f x f f x f
x x x
i e f f
+ +
− −
+ −
→ →
→ →
→ → →
− − −
= =
− −
− − − −
= = −
− −
− − −
≠ ∴
− − −
′ ∴
∵
ifferentiable at = 1
There is a corner on the graph of at (1, 0) [ the curve is not smooth around (1, 0) ] is not differentiable at 1
x f
f
Higher Derivatives (高階導數)
y = f x( )
0
( ) ( )
The first derivative, ( ) lim
h
f x h f x
f x → h
+ −
′ =
( )
0
( ) ( )
The second derivative, ( ) ( ) lim
h
f x h f x
f x f x
h
→
′ + − ′
′′ = ′ ′ =
一階導函數
二階導函數
( )
( )
0
( 1) ( 1)
( ) ( 1)
0
( ) ( )
The third derivative, ( ) ( ) lim
( ) ( )
The th derivative, ( ) ( ) lim
h
n n
n n
h
f x h f x
f x f x
h
f x h f x
n f x f x
h
→
− −
−
→
′′ + − ′′
′′′ = ′′ ′ =
+ −
= ′ =
⋮
三階導函數
Notations
( ) ( ) ( )
( )
The first derivative, d dy d df x
y y f x f x Df x
dx dx dx dx
′= = = ′ = = =
2 3 3
3
2 3 3
The third derivative, d d d d y
y y y D y
dx dx dx dx
′′′ = = = =
( ) ( ) ( ) ( )
( )
2 2
2
2 2
2 2
2
2 2
The second derivative,
d d d d y
y y y D y
dx dx dx dx
d f x
d d d
f x f x f x D f x
dx dx dx dx
′′ = = = =
′′ = = = =
( ) ( ) ( ) ( )
( )
( ) ( )
( ) ( )
2 3 3
2 3 3
3
2 3 3
4 4 4 4
4 4
4 4 4
The third derivative, The fourth derivative,
y y y D y
dx dx dx dx
d f x
d d d
f x f x f x D f x
dx dx dx dx
d d y d d
y y f x f x
dx dx dx
= = = =
′′′ = = = =
= = = = =
( )
( )
( ) ( )
( ) ( ) ( )
( )
4 4
The th derivative,
n n n n
n n n
n n n n
f x D f x
dx d f x
d d y d
n y y f x f x D f x
dx dx dx dx
=
= = = = = =
⋮
