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Slopes and the Derivative

Mathematics 53

Institute of Mathematics (UP Diliman)

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For today

1 The Tangent Line

2 Definition of the Derivative

3 Differentiability

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For today

1 The Tangent Line

2 Definition of the Derivative

3 Differentiability

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Tangent Line to a Circle

Tangent line to a circle at point P

P

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Tangent Line to a Circle

Tangent line to a circle at point P

P

How about tangent line to a curve?

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Secant Line

y =f(x)

P

P(x0, f(x0)) - point on the graph off

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Secant Line

y =f(x)

P

Q

P(x0, f(x0)) - point on the graph off

Q(x1, f(x1)) - another pt on the graph off

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Secant Line

y =f(x)

P

Q

P(x0, f(x0))

Q(x1, f(x1))

P Q is a secant line (i.e. a line thatpasses through any two points on acurve).

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Secant Line

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

P(x0, f(x0))

Q(x1, f(x1))

P Q is a secant line (i.e. a line thatpasses through any two points on acurve).

What is the slope ofP Q?

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Secant Line

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

P(x0, f(x0))

Q(x1, f(x1))

P Q is a secant line (i.e. a line thatpasses through any two points on acurve).

mPQ

= f(x1) f(x0)

x1 x0

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Secant Line

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

P(x0, f(x0))

Q(x1, f(x1))

P Q is a secant line (i.e. a line thatpasses through any two points on acurve).

mPQ

= f(x1) f(x0)

x1 x0Let x= x1 x0

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Secant Line

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

P(x0, f(x0))

Q(x1, f(x1))

P Q is a secant line (i.e. a line thatpasses through any two points on acurve).

mPQ

= f(x1) f(x0)

x1 x0Let x= x1 x0

mPQ

= f(x0+ x) f(x0)

x

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The Tangent Line to the graph off at P

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

Let Q get closer and closer to P,i.e.

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The Tangent Line to the graph off at P

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

Let Q get closer and closer to P,i.e.

Q

P

x1 x0x 0

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The Tangent Line to the graph off at P

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

Let Q get closer and closer to P,i.e.

Q

P

x1 x0x 0

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The Tangent Line to the graph off at P

y =f(x)

P

Q

f(x0)

f(x1)

x0 x1

Let Q get closer and closer to P,i.e.

Q

P

x1 x0x 0

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h h h f f P

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The Tangent Line to the graph off at P

y =f(x)

P

Q

f(x0)

f(x1)

x0x1

Let Q get closer and closer to P,i.e.

Q

P

x1 x0x 0

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Th T Li h h f f P

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The Tangent Line to the graph off at P

y =f(x)

P

Let Q get closer and closer to P,i.e.

Q

P

x1 x0x 0

The lineP Q will approach the line .

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Th T li h h f f ( f( ))

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The Tangent line to the graph off at(x0, f(x0))

y =f(x)

P

Slope of the line

.

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Th T t li t th h f f t ( f( ))

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The Tangent line to the graph off at(x0, f(x0))

y =f(x)

P

Slope of the line

m=

.

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Th T t li t th h f f t ( f( ))

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The Tangent line to the graph off at(x0, f(x0))

y =f(x)

P

Slope of the line

m= limx1x0

f(x1) f(x0)x1 x0

.

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Th T t li t th h f f t ( f( ))

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The Tangent line to the graph off at(x0, f(x0))

y =f(x)

P

Slope of the line

m= limx1x0

f(x1) f(x0)x1 x0

Recall: x=x1 x0

.

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The Tangent line to the graph of f at (x f(x ))

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The Tangent line to the graph off at(x0, f(x0))

y =f(x)

P

Slope of the line

m= limx1x0

f(x1) f(x0)x1 x0

Recall: x=x1 x0

m=

.

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The Tangent line to the graph of f at (x f(x ))

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The Tangent line to the graph off at(x0, f(x0))

y =f(x)

P

Slope of the line

m= limx1x0

f(x1) f(x0)x1 x0

Recall: x=x1 x0

m= limx0

f(x0+ x) f(x0)x

.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 7 / 34

The Tangent line to the graph of f at (x0 f(x0))

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The Tangent line to the graph off at(x0, f(x0))

y =f(x)

P

Slope of the line

m= limx1x0

f(x1) f(x0)x1 x0

Recall: x=x1 x0

m= limx0

f(x0+ x) f(x0)x

The line is called the tangent line to the graph ofy=f(x) at P.

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The Tangent Line

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The Tangent Line

Definition

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The Tangent Line

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The Tangent Line

Definition

If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line

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The Tangent Line

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The Tangent Line

Definition

If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line

1 passing through Pwhose slope is given by

m= limx0

f(x0+ x)

f(x0)

x ,provided that this limit exists.

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The Tangent Line

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The Tangent Line

Definition

If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line

1 passing through Pwhose slope is given by

m= limx0

f(x0+ x)

f(x0)

x ,provided that this limit exists.

2 with equation x=x0 if

lim

x0

f(x0+ x) f(x0)

x

= +

or

and

limx0+

f(x0+ x) f(x0)x

= + or

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The Tangent Line

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The Tangent Line

Definition

If the function f is defined at x= x0, then the tangent line to the graphoffat the point P(x0, f(x0)) is the line

1 passing through Pwhose slope is given by

m= limx0

f(x0+ x)

f(x0)

x ,

provided that this limit exists.

2 with equation x=x0 if

lim

x0

f(x0+ x) f(x0)

x

= +

or

and

limx0+

f(x0+ x) f(x0)x

= + orOtherwise, there is no tangent line to the graph off at P.

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The Tangent Line

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The Tangent Line

Remarks

1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.

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The Tangent Line

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g

Remarks

1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.

0

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The Tangent Line

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g

Remarks

1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.

2 The tangent line to the graph of a function may intersect the graph at

points other than the point of tangency.

0

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The Tangent Line

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g

Remarks

1 The slope of the tangent line to the graph off at Pgives us an ideaon the flatness or steepness of the graph off at Pand whetherthe graph offrises or falls at P.

2 The tangent line to the graph of a function may intersect the graph at

points other than the point of tangency.

0

P

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The Normal Line

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Definition

The line normal to the graph offat the point P is the line perpendicularto the tangent line at P.

P

0

NL

TL

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E l

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.Solution.

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Example

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.Solution.

Recall: Point-Slope Form of a Line : y y0=m(x x0)

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Example

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.Solution.

Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)

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Example

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.Solution.

Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m

TL

= slope of the tangent line to the graph off(x) at (1, 1)

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Example

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.Solution.

Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m

TL

= slope of the tangent line to the graph off(x) at (1, 1)

mTL

= limx0

f(1 + x) f(1)x

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Example

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.Solution.

Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m

TL

= slope of the tangent line to the graph off(x) at (1, 1)

mTL

= limx0

f(1 + x) f(1)x

= limx0

11+x 1

x

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Example

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Example

Give equations of the tangent and normal lines to the graph off(x) = 1

x

at x= 1.Solution.

Recall: Point-Slope Form of a Line : y y0=m(x x0)f(1) = 1 so the point is (1, 1)m

TL

= slope of the tangent line to the graph off(x) at (1, 1)

mTL

= limx0

f(1 + x) f(1)x

= limx0

11+x 1

x

= limx0

1 (1 + x)x(1 + x)

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= limx0

xx(1 + x)

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= limx0

xx(1 + x)

= limx0

11 + x

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= limx0

xx(1 + x)

= limx0

11 + x

mTL

=

1

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= limx0

xx(1 + x)

= limx0

11 + x

mTL

=

1

Tangent Line: y 1 = (x 1)

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= limx0

xx(1 + x)

= limx0

11 + x

mTL

=

1

Tangent Line: y 1 = (x 1)

mTL

= 1, so the slope of the normal line is 1.

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= limx0

xx(1 + x)

= limx0

11 + x

mTL

=

1

Tangent Line: y 1 = (x 1)

mTL

= 1, so the slope of the normal line is 1.

Normal Line: y 1 = (x 1)

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Example

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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.

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Example

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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.

Solution. Atx= 1 :

mTL

= limx0

g(1 + x) g(1)x

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Example

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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.

Solution. Atx= 1 :

mTL

= limx0

g(1 + x) g(1)x

= limx0

[(1 + x)2 + 2]

(12 + 2)

x

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Example

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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.

Solution. Atx= 1 :

mTL

= limx0

g(1 + x) g(1)x

= limx0

[(1 + x)2 + 2]

(12 + 2)

x

= limx0

2x+ (x)2

x

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Example

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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.

Solution. Atx= 1 :

mTL

= limx0

g(1 + x) g(1)x

= limx0

[(1 + x)2 + 2]

(12 + 2)

x

= limx0

2x+ (x)2

x

= limx0

x(2 + x)

x

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Example

2

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Find the slope of the tangent line to the graph ofg(x) =x2 + 2 at x= 1and at x= 2.

Solution. Atx= 1 :

mTL

= limx0

g(1 + x) g(1)x

= limx0

[(1 + x)2 + 2]

(12 + 2)

x

= limx0

2x+ (x)2

x

= limx0

x(2 + x)

x= lim

x0(2 + x)

= 2.

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Atx= 2 :

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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34

Atx= 2 :

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mTL

= limx0

g(2 + x) g(2)x

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Atx= 2 :

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mTL

= limx0

g(2 + x) g(2)x

= limx0

[(2 + x)2 + 2] (22 + 2)x

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Atx= 2 :

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mTL

= limx0

g(2 + x) g(2)x

= limx0

[(2 + x)2 + 2] (22 + 2)x

= limx0

[4 + 4x+ (x)2 + 2] 6x

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Atx= 2 :

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mTL

= limx0

g(2 + x) g(2)x

= limx0

[(2 + x)2 + 2] (22 + 2)x

= limx0

[4 + 4x+ (x)2 + 2] 6x

= limx0

4x+ (x)2x

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Atx= 2 :

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mTL

= limx0

g(2 + x) g(2)x

= limx0

[(2 + x)2 + 2] (22 + 2)x

= limx0

[4 + 4x+ (x)2 + 2] 6x

= limx0

4x+ (x)2x

= limx0

x(4 + x)

x

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34

Atx= 2 :

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mTL

= limx0

g(2 + x) g(2)x

= limx0

[(2 + x)2 + 2] (22 + 2)x

= limx0

[4 + 4x+ (x)2 + 2] 6x

= limx0

4x+ (x)2x

= limx0

x(4 + x)

x= lim

x0(4 + x)

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34

Atx= 2 :

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mTL

= limx0

g(2 + x) g(2)x

= limx0

[(2 + x)2 + 2] (22 + 2)x

= limx0

[4 + 4x+ (x)2 + 2] 6x

= limx0

4x+ (x)2x

= limx0

x(4 + x)

x= lim

x0

(4 + x)

= 4.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 14 / 34

As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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( )

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34

As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL = limx0

g(x+ x)

g(x)

x

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34

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mTL = limx0

g(x+ x)

g(x)

x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

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As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x)

g(x)

x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= lim

x

0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x

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As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x)

g(x)

x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= lim

x

0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x= lim

x0

2xx+ (x)2

x

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As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x)

g(x)

x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= lim

x

0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x= lim

x0

2xx+ (x)2

x= lim

x0(2x+ x)

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As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x) g(x)x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= limx0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x= lim

x0

2xx+ (x)2

x= lim

x0(2x+ x)

= 2x.

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As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x) g(x)x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= limx0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x= lim

x0

2xx+ (x)2

x= lim

x0(2x+ x)

= 2x.

Atx= 1:

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As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x) g(x)x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= limx0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x= lim

x0

2xx+ (x)2

x= lim

x0(2x+ x)

= 2x.

Atx= 1: mTL= 2 1 = 2

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As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x) g(x)x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= limx0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x

= limx0

2xx+ (x)2

x= lim

x0(2x+ x)

= 2x.

Atx= 1: mTL= 2 1 = 2Atx= 2:

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34

As a matter of fact, we can compute for the slope of the tangent line tog(x) for any value ofx.

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mTL

= limx0

g(x+ x) g(x)x

= limx0

[(x+ x)2 + 2] (x2 + 2)x

= limx0

[x2 + 2xx+ (x)2 + 2] (x2 + 2)

x

= limx0

2xx+ (x)2

x= lim

x0(2x+ x)

= 2x.

Atx= 1: mTL= 2 1 = 2Atx= 2: mTL= 2 2 = 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 15 / 34

For today

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1 The Tangent Line

2 Definition of the Derivative

3 Differentiability

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Definition of the Derivative

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In the previous example, we computed a function that gives the slope ofthe tangent line to the graph of a function f(x) at any value ofx. Thisfunction is actually called the derivative off(x).

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Definition of the Derivative

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In the previous example, we computed a function that gives the slope ofthe tangent line to the graph of a function f(x) at any value ofx. Thisfunction is actually called the derivative off(x).

Definition

The derivative of a function f(x), denoted by f

(x), is the function

f(x) = limx0

f(x+ x) f(x)x

.

It is defined at all pointsx

in the domain off

where the limit exists.

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Definition of the Derivative

Remarks

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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 18 / 34

Definition of the Derivative

Remarks

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1

There may be points xo dom fat which f

(x0) does not exist. So,dom f dom f.

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Definition of the Derivative

Remarks

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1

There may be points xo dom fat which f

(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to

the graph of the function at point P(x0, f(x0)).

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Definition of the Derivative

Remarks

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1

There may be points xo dom fat which f

(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to

the graph of the function at point P(x0, f(x0)).

3 To get the derivative off at x=x0, we use

f(x0) = limx0

f(x0+ x) f(x0)x

.

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Definition of the Derivative

Remarks

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1

There may be points xo dom fat which f

(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to

the graph of the function at point P(x0, f(x0)).

3 To get the derivative off at x=x0, we use

f(x0) = limx0

f(x0+ x) f(x0)x

.

Alternatively,

f(x0) = limxx0

f(x) f(x0)x x0

.

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Definition of the Derivative

Remarks

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1

There may be points xo dom fat which f

(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to

the graph of the function at point P(x0, f(x0)).

3 To get the derivative off at x=x0, we use

f(x0) = limx0

f(x0+ x) f(x0)x

.

Alternatively,

f(x0) = limxx0

f(x) f(x0)x x0

.

4 Other notations: y ify=f(x), dy

dx ,

d

dx[f(x)] , Dx[f(x)].

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Definition of the Derivative

Remarks

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1

There may be points xo dom fat which f

(x0) does not exist. So,dom f dom f.2 The definition tells us that f(x0) is the slope of the tangent line to

the graph of the function at point P(x0, f(x0)).

3 To get the derivative off at x=x0, we use

f(x0) = limx0

f(x0+ x) f(x0)x

.

Alternatively,

f(x0) = limxx0

f(x) f(x0)x x0

.

4 Other notations: y ify=f(x), dy

dx ,

d

dx[f(x)] , Dx[f(x)].

5 The process of computing the derivative is called differentiation.

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Example

Find the derivative off(x) = x.

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Example

Find the derivative off(x) = x.

Solution. Using the definition of the derivative we have

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Definition of the Derivative

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Example

Find the derivative off(x) = x.

Solution. Using the definition of the derivative we have

f

(x) = limx0

x+ x

x

x

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Definition of the Derivative

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Example

Find the derivative off(x) = x.

Solution. Using the definition of the derivative we have

f

(x) = limx0

x+ x

x

x

= limx0

x+ xx

x

x+ x+

x

x+ x+

x

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Example

Find the derivative off(x) = x.

Solution. Using the definition of the derivative we have

f

(x) = limx0

x+ x

x

x

= limx0

x+ xx

x

x+ x+

x

x+ x+

x

= limx0

(x+ x)

x

x(x+ x+ x)

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1

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= limx0

1

x+ x+ x

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1

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= limx0

1

x+ x+ x=

1

2

x.

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1

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= limx0

1

x+ x+ x=

1

2

x.

f

(x) =

1

2x or Dx[x ] = 1

2x .

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1

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= limx0

1

x+ x+ x=

1

2

x.

f

(x) =

1

2x or Dx[x ] = 1

2x .Note:

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1

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= limx0

1

x+ x+ x=

1

2

x.

f

(x) =

1

2x or Dx[x ] = 1

2x .Note:

dom f = [0, +)

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 20 / 34

1

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= limx0

1

x+ x+ x=

1

2

x.

f

(x) =

1

2x or Dx[x ] = 1

2x .Note:

dom f = [0, +)dom f = (0, +

)

Instit te of Mathematics (UP Diliman) Slo es and the De i ati e Mathematics 53 20 / 34

Example

Find the derivative off(x) = 3

x at x= 0.

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Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.

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Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.Using the alternative definition:

f(0) = limx0

3

x 30x 0

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Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.Using the alternative definition:

f(0) = limx0

3

x 30x 0

= limx0

3

xx

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Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.Using the alternative definition:

f(0) = limx0

3

x 30x 0

= limx0

3

xx

= limx0

1

x23

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34

Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.Using the alternative definition:

f(0) = limx0

3

x 30x 0

= limx0

3

xx

= limx0

1

x23

=

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34

Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.Using the alternative definition:

f(0) = limx0

3

x 30x 0

= limx0

3

xx

= limx0

1

x23

= 1

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34

Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.Using the alternative definition:

f(0) = limx0

3

x 30x 0

= limx0

3

xx

= limx0

1

x23

= 1

0+

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 21 / 34

Example

Find the derivative off(x) = 3

x at x= 0.

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Solution.Using the alternative definition:

f(0) = limx0

3

x 30x 0

= limx0

3

xx

= limx0

1

x23

= 1

0+

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For today

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1 The Tangent Line

2 Definition of the Derivative

3 Differentiability

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Differentiability

Definitions

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Differentiability

Definitions

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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.

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Differentiability

Definitions

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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.

2 The function f is differentiable on (a, b) iff is differentiable at

every real number in (a, b

).

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Differentiability

Definitions

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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.

2 The function f is differentiable on (a, b) iff is differentiable atevery real number in (a, b).

3 The function f is differentiable everywhere if it is differentiable atevery real number.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 23 / 34

Differentiability

Definitions

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1 If the derivative f(x0) of the function f at x= x0 exists, then f issaid to be differentiable at x= x0.

2 The function f is differentiable on (a, b) iff is differentiable atevery real number in (a, b).

3 The function f is differentiable everywhere if it is differentiable atevery real number.

RemarkThe function f(x) is differentiable at x= x0 if and only ifx0dom f.

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Example

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ExampleWe know that Dx[

x ] =

1

2

x.

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Example

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Example

We know that Dx[

x ] = 1

2

x.

= f(x) = x is differentiable at any positive real number x

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Example

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Example

We know that Dx[

x ] = 1

2

x.

= f(x) = x is differentiable at any positive real number x

= f is not differentiable at x= 0.

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Differentiability

Definitions

Let the function f(x) be defined at x= x0.

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f ( )

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Differentiability

Definitions

Let the function f(x) be defined at x= x0.

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1 The derivative from the left off(x) at x=x0, denoted by f

(x0),

is given by

f

(x0) = lim

xx

0

f(x) f(x0)

x x0.

2 The derivative from the right off(x) at x= x0, denoted byf+(x0), is given by

f+(x0) = limxx+0

f(x) f(x0)x x0 .

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Differentiability

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Remark

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Differentiability

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Remark

1 The derivative from the left [right] is also referred to as the left-handderivative [right-hand derivative], or simply left derivative [rightderivative].

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Differentiability

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Remark

1 The derivative from the left [right] is also referred to as the left-handderivative [right-hand derivative], or simply left derivative [rightderivative].

2 The function f is differentiable at x=x0 if and only iff

(x0) and

f+(x0) exist and

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Differentiability

k

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Remark

1 The derivative from the left [right] is also referred to as the left-handderivative [right-hand derivative], or simply left derivative [rightderivative].

2 The function f is differentiable at x=x0 if and only iff

(x0) and

f+(x0) exist and

f

(x0) =f

+(x0) =f(x0).

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution.

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0)

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

= 1

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

= 1

f+(0)

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34

Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

= 1

f+(0) = limx0+

f(x) f(0)x

0

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Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

= 1

f+(0) = limx0+

f(x) f(0)x

0

= limx0+

x 0x

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34

Example

Determine iff(x) = |x| =

x , x 0x , x

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Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

= 1

f+(0) = limx0+

f(x) f(0)x

0

= limx0+

x 0x

= 1

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34

Example

Determine iff(x) = |x| =

x , x 0x , x

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135/177

Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

= 1

f+(0) = limx0+

f(x) f(0)x

0

= limx0+

x 0x

= 1

=

f

(0)

=f+(0)

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34

Example

Determine iff(x) = |x| =

x , x 0x , x

8/13/2019 Slopes and the Derivative

136/177

Solution. We compute for the derivatives from the left and from the rightat x= 0.

f

(0) = limx0

f(x) f(0)x

0

= limx0

x 0x

= 1

f+(0) = limx0+

f(x) f(0)x

0

= limx0+

x 0x

= 1

=

f

(0)

=f+(0)

f(x) is not differentiable at x= 0.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 27 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

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138/177

We compute for the derivatives from the left and from the right at x= 4.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

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We compute for the derivatives from the left and from the right at x= 4.

f

(4)

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

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We compute for the derivatives from the left and from the right at x= 4.

f

(4) = limx4

( 12

x2 + 24) 32x 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

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141/177

We compute for the derivatives from the left and from the right at x 4.

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2x2

8x 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

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We compute for the derivatives from the left and from the right at x 4.

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

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p g

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

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p g

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4)

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

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p g

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4) = limx4+

16

x 32x 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

146/177

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4) = limx4+

16

x 32x 4

= limx4+ 16

x

2

x 4

x+ 2x+ 2

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

147/177

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4) = limx4+

16

x 32x 4

= limx4+ 16

x

2

x 4

x+ 2x+ 2

= limx4+

16 x 4

(x 4)x+ 2

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

148/177

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4) = limx4+

16

x 32x 4

= limx4+ 16

x

2

x 4

x+ 2x+ 2

= limx4+

16 x 4

(x 4)x+ 2= 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

149/177

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4) = limx4+

16

x 32x 4

= limx4+ 16

x

2

x 4

x+ 2x+ 2

= limx4+

16 x 4

(x 4)x+ 2= 4

f

(4) =f+(4) = 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

150/177

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4) = limx4+

16

x 32x 4

= limx4+ 16

x

2

x 4

x+ 2x+ 2

= limx4+

16 x 4

(x 4)x+ 2= 4

f

(4) =f+(4) = 4 = f(x) is differentiable at x= 4

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Example

Determine iff(x) =

12x

2 + 24 , x

8/13/2019 Slopes and the Derivative

151/177

f

(4) = limx4

( 12

x2 + 24) 32x 4

= limx4

1

2

x2

8

x 4= lim

x4

1

2

(x+ 4)(x 4)x 4

= 4

f+(4) = limx4+

16

x 32x 4

= limx4+ 16

x

2

x 4

x+ 2x+ 2

= limx4+

16 x 4

(x 4)x+ 2= 4

f

(4) =f+(4) = 4 = f(x) is differentiable at x= 4 and f(4) = 4.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 28 / 34

Differentiability and Continuity

http://goforward/http://find/http://goback/

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152/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34

Differentiability and Continuity

Theorem

Iff is differentiable atx= x0, then f is continuous atx=x0.

http://find/

8/13/2019 Slopes and the Derivative

153/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34

Differentiability and Continuity

Theorem

Iff is differentiable atx= x0, then f is continuous atx=x0.

http://find/

8/13/2019 Slopes and the Derivative

154/177

Remarks

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34

Differentiability and Continuity

Theorem

Iff is differentiable atx= x0, then f is continuous atx=x0.

http://find/

8/13/2019 Slopes and the Derivative

155/177

Remarks

Iff is discontinuous at x= x0, then f is not differentiable at x=x0.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34

Differentiability and Continuity

Theorem

Iff is differentiable atx= x0, then f is continuous atx=x0.

http://find/

8/13/2019 Slopes and the Derivative

156/177

Remarks

Iff is discontinuous at x= x0, then f is not differentiable at x=x0.

Iff is continuous at x= x0, it does not mean that f is differentiableat x=x0.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34

Differentiability and Continuity

Theorem

Iff is differentiable atx= x0, then f is continuous atx=x0.

http://find/

8/13/2019 Slopes and the Derivative

157/177

Remarks

Iff is discontinuous at x= x0, then f is not differentiable at x=x0.

Iff is continuous at x= x0, it does not mean that f is differentiableat x=x0.

Iff is not differentiable at x= x0, it does not mean that f is notcontinuous at x= x0.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 29 / 34

Example

Consider the function f(x) = |x|.

http://find/

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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 30 / 34

Example

Consider the function f(x) = |x|.

http://find/

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f is continuous at x= 0

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 30 / 34

Example

Consider the function f(x) = |x|.

http://find/

8/13/2019 Slopes and the Derivative

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f is continuous at x= 0

f is not differentiable at x= 0

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 30 / 34

http://find/

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161/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34

Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.

http://find/

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162/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34

Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.

Remarks

http://find/

8/13/2019 Slopes and the Derivative

163/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34

Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.

Remarks

A function f is not differentiable at x=x0 if one of the following is true:

http://find/

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164/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34

Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.

Remarks

A function f is not differentiable at x=x0 if one of the following is true:f is discontinuous at x= x0

http://find/

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165/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34

Geometrically, iff is differentiable at x=x0, then the graph off hastangent line at x=x0.

Remarks

A function f is not differentiable at x=x0 if one of the following is true:f is discontinuous at x= x0

http://find/

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x0

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 31 / 34

Remarks

the graph offhas a vertical tangent line at x=x0

http://find/

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167/177

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34

Remarks

the graph offhas a vertical tangent line at x=x0

http://find/

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x0

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34

Remarks

the graph offhas a vertical tangent line at x=x0

the graph offhas no well-defined tangent line at x=x0, i.e., the

graph offhas a corner or cusp at x= x0.

http://find/

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x0

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34

Remarks

the graph offhas a vertical tangent line at x=x0

the graph offhas no well-defined tangent line at x=x0, i.e., the

graph offhas a corner or cusp at x= x0.

http://find/

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x0 x0

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 32 / 34

* * * The End * * *

http://find/

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171/177

Next Meeting:

Differentiation Rules

The Chain Rule

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 33 / 34

Exercise

http://find/

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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34

Exercise

Determine if

f(x) =

x2 , x < 1

1 2x , x 1is differentiable at x0= 1.

http://find/

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Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34

Exercise

Determine if

f(x) =

x2 , x < 1

1 2x , x 1is differentiable at x0= 1.Solution.

http://find/

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Solution.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34

Exercise

Determine if

f(x) =

x2 , x < 1

1 2x , x 1is differentiable at x0= 1.Solution.

http://find/

8/13/2019 Slopes and the Derivative

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f

(1) = limx1

x2 1x+ 1

= limx1

(x+ 1)(x 1)x+ 1

= 2

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34

Exercise

Determine if

f(x) =

x2 , x < 1

1 2x , x 1is differentiable at x0= 1.Solution.

http://find/

8/13/2019 Slopes and the Derivative

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f

(1) = limx1

x2 1x+ 1

= limx1

(x+ 1)(x 1)x+ 1

= 2

f+(

1) = lim

x3

+

(1 2x) 1

x+ 1= lim

x3+

2(x+ 1)x+ 1

= 2

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34

Exercise

Determine if

f(x) =

x2 , x < 1

1 2x , x 1is differentiable at x0= 1.Solution.

http://find/

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f

(1) = limx1

x2 1x+ 1

= limx1

(x+ 1)(x 1)x+ 1

= 2

f+(

1) = lim

x3

+

(1 2x) 1x

+ 1= lim

x3+

2(x+ 1)x+ 1

= 2

f

(1) =f

+(1) = 2f(x) is differentiable at x= 1 and f(1) = 2.

Institute of Mathematics (UP Diliman) Slopes and the Derivative Mathematics 53 34 / 34

http://find/