Differentiation at one point

Differentiation
• Concept of Differentiation………………….2
• Derivative of Trigonometric Function…..20
• The Chain Rule……………………………...23
• Implicit Differentiation……………………..27
• Tangent Line and Normal Line…………...30

Concept of Differentiation
Differentiation at one point
Introduction ( two topic in the same theme)
a. Tangent Lines
The secant line connecting P and Q has slope
mPQ =

f ( x ) − f (c )
x−c

If x c , then the secant line
through P and Q will approach
the tangent line at P. Thus the
slope of the tangent line is

Q

f(x)

f(x)-f(c)
f(c)

P

f ( x ) − f (c )
m = lim
x →c
x−c
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x-c
c
x
2

Differentiation
chapter

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b. Instantaneous Velocity
Let a particle travel around an axis and position of
particle at time t is s = f(t). If the particle has a
coordinate f(c) at time c and f(c+h) at time c + h.
Elapsed time

distance
traveled

c

f((c)

c+h

f((c+h)
s

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Differentiation
chapter

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Thus, the average velocity during the time interval
[c,c+h] is
f (c + h) − f (c)
vrata-rata =
h
If h 0, we get instantaneous velocity at x = c:

v = lim vrata-rata
h →0

f (c + h ) − f (c )
= lim
h →0
h

Let x = c + h, instantaneous velocity can be written as

v = lim
x →c

f(x) − f(c)
x−c

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Differentiation
chapter

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From two cases: The slope of the tangent line and
instantaneous velocity has the same
formula.
Definition : The first derivative of f at x = c,
denoted by f ′(c) is defined by

f ( x) − f (c)
f ′(c) = lim
x→c
x −c
if its limit exist.

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Differentiation
chapter

Other notation :

4

df (c)
, y′(c)
dx

Example: Let f ( x ) =

1
evaluate f ' (3)
x

1 1
−

f(x) − f( 3 )
= lim x 3
f'( 3 ) = lim
x→3
x→3 x − 3
x−3
3− x
− ( x − 3)
= lim
= lim
x → 3 3 x ( x − 3)
x → 3 3 x ( x − 3)
−1
1
= lim
= −
x→ 3 3 x
9
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Derivative from the right
and the left
• Derivatives from the left at c, denoted by f−′ (c) , is
defined by
f ( x ) − f (c )
f −′ (c) = lim−
x →c
x−c
• Derivatives from the right at c, denoted by f+′ (c) , is
defined by
f ( x ) − f (c )
′
f + (c) = lim+
x →c
x−c
• A function f is said differentiable at c ( f ′(c ) exist) if
f −′ (c) = f +′ (c)

and
f ′(c) = f −′ (c) = f +′ (c)
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Example
Let f ( x) =

x2 − x + 3 , x < 1
1+ 2 x

, x ≥1

Determine whether f (x) is differentiable at x = 1.
'
f
If f is differentiable find (1) .
Answer:
a.

2
f
(
x
)
−
f
(1)
x
− x +3−(1+ 2 1)
′
f− (1) = lim−
= lim−
x→1
x→1
x −1
x −1

x2 − x

x ( x − 1)
= lim−
= lim−
=1
x →1
x →1
x −1
x −1
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Differentiation
chapter

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f ( x) − f (1)
1 + 2 x − (1 + 2 1)
′
b. f + (1) = lim+
= lim+
x →1
x →1
x −1
x −1
2 x −2
x −1
= lim+
= 2 lim+
=1
x →1
x →1 (
x −1
x − 1)( x + 1)
Thus, f is differentiable at x = 1 f ′(1) = 1 .

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9

Differentiable
and Continuity
Theorem: If f is differentiable at point c then f is
continuous at c.
f ( x ) = f (c )
Proof: We will prove xlim
→c
f ( x ) − f (c )
.( x − c ) , x ≠ c
Since
x−c
f ( x ) − f (c )
( x − c)
Thus, lim f ( x) = lim f (c) +
x →c
x →c
x−c
f ( x ) = f (c ) +

f ( x ) − f (c )
= lim f (c) + lim
⋅ lim( x − c) = f (c ) + f '(c) ⋅ 0 = f (c )
x →c
x →c
x →c
x−c

The converse, however, is false , a function may be
continuous at a point but not differentiable.
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Example
1. Show that, f (x) = | x | is continuous at x = 0 but f (x)
is not differentiable at x = 0.
Answer: We will show f (x) = | x | is continuous at x = 0
f ( x) =| x |=

x , x≥0
−x , x