Benginning Calculus Lecture notes 3 - derivatives

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.




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)




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




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)




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 .




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).




Example

Differentiate y = x2 − 8x + 9 at x = 2.




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




Example

Let f (x) = x3 − x . The derivative of f is:




Example

Let f (x) =√x . Then, f ′ (x) is:




Example

Let f (x) =1− x2+ x

. Then, f ′ (x) is:




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




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.




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.




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




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.




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




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.




Differentiable Implies Continuous

Theorem 2

If f (x) is a differentiable at a, then f (x) is continuous at a.




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)



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Benginning Calculus Lecture notes 3 - derivatives