MATH 136
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The Formal Limit Definition
of Derivative
Given the graph of y = f (x) , we wish to derive the formula for the slope of the tangent
line when x = a . To do so, we first consider the slope of the line through the two points
f (x) − f (a)
( a , f (a) ) and ( x , f (x) ). This line has slope m =
. In order to obtain the slope
x −a
of the tangent line at the point a , we must allow x to move closer and closer to a ; thus,
€ must take a limit. The slope of the tangent line at the point a is then defined to be
we
€
f ( x) − f (a)
(1)
f ′(a) = lim
€
€
€
x−a €
x →a
Once we have this formula for f ′(a) , we can substitute x for a to obtain a general
formula for the derivative function f ′ (x ) . Alternately, we can use the direct form
f (x + h) − f (x )
.€
€
h
h→ 0
(2)
f ′ (x ) = lim
This version results from considering the slope between the two points ( x , f (x) ) and
( x + h , f ( x + h) ), then letting h become smaller and smaller.
€
Example 1. Let f (x) =
€
€
x . Use the formal limit definition of derivative to derive f ′(x) .
Solution. We first use Definition (1) to derive the derivative at point a :
f (x ) − f (a)
= lim
x −a
x→ a
x →a
f ′ (a) = lim
x− a
x − a  x + a
= lim


x−a
x → a €x − a  x + a 
x −a
= lim
x → a (x − a)( x + a )
x →a
= lim
Thus, f ′(x) =
1
=
x+ a
1
1
=
.
a+ a 2 a
1
, which is defined for all x > 0.
2 x
We note that f (x) = x is defined for all x ≥ 0 , but f ′(x) is not defined at the
endpoint x = 0 . Because lim f ′( x) = + ∞ , the graph of x is becoming vertical as x
€
x →0+
decreases to 0.
€
€
€ Alternately using (2), we have
f (x + h) − f (x )
= lim
h
h→ 0
h→ 0
f ′ (x ) = lim
x+h − x
x+h− x
= lim
×
h
h
h→ 0
x +h − x
1
1
= lim
=
, for x > 0.
2 x
h → 0 h( x + h + x )
h→ 0 x + h + x
= lim
x+h+ x
x+h+ x
Example 2. Let f (x) =
f ′(2) .
20
. Use the formal limit definition of derivative to evaluate
x2
Solution. Using (1) with a = 2 , we have
20
2 −5
f ( x) − f (2)
20 − 5x 2
5(2 − x)(2 + x)
x
f ′ (2) = lim
=
lim
=
lim
= lim
€
2
x−2
x→ 2
x→ 2 x − 2
x → 2 x (x − 2)
x→ 2
x 2 (x − 2)
−5(2 + x) −20
=
= −5 .
4
x→ 2
x2
= lim
Alternately using (2), we have
20
20
− 2
2
f (x + h) − f (x )
x = lim 20 x 2 − 20( x + h)2
f ′ (x ) = lim
= lim ( x + h)
h→ 0
h
h→ 0
h
h→0
x 2 (x + h)2 h
20x 2 − 20(x 2 + 2x h + h2 )
−20(2 x h + h2 )
−20(2x + h)
=
lim
= lim 2
2
2
2
2
2
h→ 0
x ( x + h) h
h → 0 x ( x + h) h
h → 0 x (x + h)
= lim
=
−40x −40
= 3 , for x ≠ 0.
x4
x
Thus, f ′(2) = –40/8 = –5.
Example 3. Let f (x) = 4 x 3. Use the formal limit definition to evaluate f ′(−3) .
Solution. Using (1) with a = −3 , we have
€
f ( x) − f (−3)
4x 3 − (−108)
4(x 3 + 27)
= lim
= lim
x − (−3)
x+3
x +3
x → −3
x → −3
x → −3
f ′ (−3) = lim
4( x + 3)( x 2 − 3x + 9)
= lim 4( x 2 − 3x + 9) = 4 × (9 + 9 + 9) = 108 .
x
+
3
x → −3
x → −3
= lim
Alternately using (2), we have
f (x + h) − f (x )
4( x + h)3 − 4x 3
4(x 3 + 3x 2 h + 3xh 2 + h3 ) − 4 x 3
= lim
= lim
h
h
h
h→ 0
h→ 0
h →0
f ′ (x ) = lim
4(3x 2 h + 3xh 2 + h3 )
= lim 4(3x 2 + 3xh + h 2 ) = 12 x 2 , for all x.
h
h→ 0
h→0
= lim
So f ′(−3) = 12(−3)2 = 108 .
2
Example 4. Let f (x) = 10x −
evaluate f ′(4) .
8
. Use the formal limit definition of derivative to
x
Solution. Using (1) with a = 4 , we have

8
8

 10x 2 −  − (160 − 2 )
10x 2 − 160 −  − 2
f (x ) − f (4)


x
x
f ′ (4) = lim €
= lim
= lim
x −4
x−4
x−4
x→ 4
x →4
x→ 4
(
10( x − 4)(x + 4) −
= lim
x−4
x→ 4
)
2(4 − x)
2
1
x
= lim 10( x + 4) + lim
= 80 + = 80. 5 .
x→ 4
x→ 4 x
2
Note: The key in this example is to group the terms separately for the two parts of the
function before factoring and separating the limits. In other words, apply the limit of a
sum/difference as being the sum/difference of the limits.)
Derivatives of the Trigonometric Functions
We now shall use many of our previous results to derive the derivatives of sin x and
cos x . Recall the following facts:
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B − sin A sin B
€
sin x
cos x − 1
lim
= 1 and lim
=0
x
x →0 x
x →0
€
€
Now using the derivative form in (2), we obtain
d(sin x)
sin(x + h) − sin x
sin x cos h + cos x sinh − sin x
= lim
= lim
dx
h
h
h→0
h→0
sin x (cosh −1) + cos x sinh
sin x (cosh −1)
cos x sin h
= lim
= lim
+ lim
h
h
h
h→0
h→0
h→0
cosh −1
sin h
+ cos x lim
= sin x × 0 + cos x ×1 = cos x .
h
h→0
h→0 h
= sin x lim
To obtain the derivative of cos x , first note that the cosine graph is simply the sine
€ graph shifted to left by π/2. Thus, cos x = sin( x + π / 2) . However if we shift the cosine
graph to the left by π/2, we obtain a negative sine graph. Thus, cos(x + π / 2) = − sin x .
Now we have
€
d (cos x) d(sin( x + π / 2))
=
= cos( x + π / 2) = − sin x .
dx
dx
We also can derive the derivatives of tan x and sec x using the Quotient and Chain
Rules:
 sin x 
d (sin x )
d ( cos x )


x − sin x ×
€ × cos€
d (tan x) d  cos x 
dx
dx
=
=
2
dx
dx
cos x
=
cos x × cos x − sin x × (− sin x) cos 2 x + sin 2 x
1
2
=
=
2
2
2 = sec x
cos x
cos x
cos x
and
(
)
−1
d (sec x) d (cos x)
d(cos x)
=
= −(cos x )−2 ×
dx
dx
dx
= −(cos x )−2 × − sin x =
sin x
1
sin x
2 = cos x × cos x = sec x tan x .
cos x