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Universiti Pendidikan Sultan Idris
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Beginning Calculus- The Derivatives -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,FSMT - UPSI
(LECTURE SLIDES SERIES)
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 1 / 21
The Derivatives of Functions Differentiability
Learning Outcomes
Compute the slopes of secant and tangent lines.
Evaluate the derivative of functions using limits.
Determine the differentiability of a function.
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 2 / 21
The Derivatives of Functions Differentiability
Tangent and Secant Lines
y
x
f(x0)
x0
P(x0, y0) Tangent LineSecant
Line
y = f(x)Q
)( 0 xxf ∆+
y∆
x∆
xx ∆+0
Slope of the secant line:
mPQ =change in ychange in x
=∆y∆x
=f (x0 + ∆x)− f (x0)
∆x(1)
Q −→ P; ∆x −→ 0; secant line−→tangent line.Slope of the tangent line at P (x0, y0):
mtan = lim∆x→0
∆y∆x
= lim∆x→0
f (x0 + ∆x)− f (x0)∆x
(2)
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 3 / 21
The Derivatives of Functions Differentiability
Example
The equation of the tangent line to the parabola f (x) = x2 at thepoint (1, 1) is
y − y0 = mtan (x − x0)
mtan = lim∆x→0
f (1+ ∆x)− f (1)∆x
= lim∆x→0
(1+ ∆x)2 − 1∆x
= lim∆x→0
∆x2 + 2∆x∆x
= lim∆x→0
∆x (∆x + 2)∆x
= lim∆x→0
(∆x + 2) = 2
So, the equation of the tangent line is
y − 1 = 2 (x − 1)y = 2x − 1
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 4 / 21
The Derivatives of Functions Differentiability
Rate of Change
The difference quotient
4y4x =
f (x0 + ∆x)− f (x0)∆x
(3)
is called the average rate of change of y with respect to x atx = x0.
The instantaneous rate of change of y with respect to x atx = x0 is
lim4x→0
4y4x = lim
4x→0f (x0 + ∆x)− f (x0)
∆x(4)
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 5 / 21
The Derivatives of Functions Differentiability
The Derivatives
The derivative of y = f (x) at x = x0 is
f ′ (x0) = lim4x→0
4y4x = lim
4x→0f (x0 + ∆x)− f (x0)
∆x(5)
provided that the limit exists.
The derivative of y = f (x) (at any x in the domain) is
f ′ (x) = lim4x→0
4y4x = lim
4x→0f (x + ∆x)− f (x)
∆x(6)
provided that the limit exists. f ′ (x) is called the derivativefunction of f .
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 6 / 21
The Derivatives of Functions Differentiability
Derivative Notations
The derivative of y = f (x) can be denoted as follows:
y ′ = f ′ (x) =dydx=dfdx=ddx(y) =
ddxf (x) (7)
The symboldydx
was introduced by Leibniz.
dydx= lim4x→0
4y4x (8)
In Leibniz notation, we use the notation
dydx
∣∣∣∣x=x0
ordydx
]x=x0
(9)
to indicate f ′ (x0).
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 7 / 21
The Derivatives of Functions Differentiability
Example
Differentiate y = x2 − 8x + 9 at x = 2.
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 8 / 21
The Derivatives of Functions Differentiability
Example
Let f (x) = x2. Then,
f ′ (x) = lim∆x→0
f (x + ∆x)− f (x)∆x
= lim∆x→0
(x + ∆x)2 − x2∆x
= lim∆x→0
x2 + 2x∆x + ∆x2 − x2∆x
= lim∆x→0
∆x (2x + ∆x)∆x
= lim∆x→0
(2x + ∆x) = 2x
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 9 / 21
The Derivatives of Functions Differentiability
Example
Let f (x) = x3 − x . The derivative of f is:
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 10 / 21
The Derivatives of Functions Differentiability
Example
Let f (x) =√x . Then, f ′ (x) is:
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 11 / 21
The Derivatives of Functions Differentiability
Example
Let f (x) =1− x2+ x
. Then, f ′ (x) is:
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 12 / 21
The Derivatives of Functions Differentiability
Example
Let y =1x. Then,
ddx
(1x
)= lim
∆x→0
1x + ∆x
− 1x
∆x
= lim∆x→0
x − x − ∆x
x2∆x + x (∆x)2
= lim∆x→0
−∆x∆x (x2 + x∆x)
= lim∆x→0
−1x2 + x∆x
= − 1x2
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 13 / 21
The Derivatives of Functions Differentiability
Differentiability
Definition 1
A function f is differentiable at x0 if f ′ (x0) exists. It is differentiableon an open interval (a, b) (or (b,∞) or (−∞, a) or (−∞,∞) ) if it isdifferentiable at every number in the interval.
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 14 / 21
The Derivatives of Functions Differentiability
Example
Where is the function f (x) = |x | differentiable?For x > 0, then |x | = x , and |x + ∆x | = x + ∆x . So,
f ′ (x) = lim∆x→0
|x + h| − |x |h
= lim∆x→0
x + ∆x − x∆x
= 1
and so f is differentiable for any x > 0. For x < 0, we have |x | = −xand |x + ∆x | = − (x + ∆x) . So,
f ′ (x) = lim∆x→0
|x + ∆x | − |x |∆x
= lim∆x→0
− (x + ∆x)− (−x)∆x
= −1
and also differentiable for any x < 0.
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 15 / 21
The Derivatives of Functions Differentiability
Example - continue
For x = 0,
f ′ (0) = lim∆x→0+
|0+ ∆x | − |0|∆x
= lim∆x→0+
|∆x |∆x
= lim∆x→0+
∆x∆x
= 1
f ′ (0) = lim∆x→0−
|0+ ∆x | − |0|∆x
= lim∆x→0−
|∆x |∆x
= lim∆x→0−
−∆x∆x
= −1
So, f ′ (0) does not exist. Thus f is differentiable at all x except at 0.
4 2 0 2 4
2
4
x
y
4 2 2 4
4
2
2
4
x
y
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 16 / 21
The Derivatives of Functions Differentiability
Functions Fails to be Differentiable
1 The graph of f has a sharp "corner" at a point2 f is not continuous at the point3 The graph of f has a vertical tangent.
y
xa0
1.
y
xa0
2.
y
xa0
3.
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 17 / 21
The Derivatives of Functions Differentiability
Example
Investigate the differentiability of f (x) = x1/3 at x = 0.
4 2 2 4
1.5
1.0
0.5
0.5
1.0
1.5
x
y
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 18 / 21
The Derivatives of Functions Differentiability
Example - continue
The function has a vertical tangent at x = 0. Thus, the function fails tohave a derivative at x = 0. We can show this algebraically:
f ′ (0) = lim∆x→0
f (0+ ∆x)− f (0)∆x
= lim∆x→0
(∆x)1/3 − 0∆x
= lim∆x→0
1
(∆x)2/3
As ∆x → 0, the denominator becomes small, so the function growswithout bound. This continuous function is differentiable everywhereexcept at x = 0.
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 19 / 21
The Derivatives of Functions Differentiability
Differentiable Implies Continuous
Theorem 2
If f (x) is a differentiable at a, then f (x) is continuous at a.
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 20 / 21
The Derivatives of Functions Differentiability
Proof:Let f (x) be a differentiable function at x = a. Then,
f ′ (a) = limx→a
f (x)− f (a)x − a
and the limit exists. Show that limx→a
f (x) = f (a) . So,
f (x)− f (a) = (x − a) · f (x)− f (a)x − a
limx→a
[f (x)− f (a)] = limx→a
[(x − a) · f (x)− f (a)
x − a
]= lim
x→a(x − a) · lim
x→af (x)− f (a)
x − a= lim
x→a(x − a) · f ′ (a)
= 0 · f ′ (a) = 0limx→a
f (x) = f (a)
VillaRINO DoMath, FSMT-UPSI
(D2) The Derivatives 21 / 21
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