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8/7/2019 Derivative Slides http://slidepdf.com/reader/full/derivative-slides 1/35 Differentiability and Derivative: If is an interval of R, then a function : R is said to be differentiable at 0 if lim 0 ()(0 ) 0 (or, equivalently lim 0 (0 +)(0 ) ) exists in R.

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Page 1: Derivative Slides

8/7/2019 Derivative Slides

http://slidepdf.com/reader/full/derivative-slides 1/35

Differentiability and Derivative: If I  is an interval of R, then a

function f  : I → R is said to be differentiable at x 0 ∈ I  if

limx →x 0

f (x )−f (x 0)x −x 0

(or, equivalently limh →0

f (x 0+h )−f (x 0)h  ) exists in R.

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Differentiability and Derivative: If I  is an interval of R, then a

function f  : I → R is said to be differentiable at x 0 ∈ I  if

limx →x 0

f (x )−f (x 0)x −x 0

(or, equivalently limh →0

f (x 0+h )−f (x 0)h  ) exists in R.

If f  is differentiable at x 0, then the derivative of f  at x 0 is

f (x 

0) = limx →x 0

f (x )

−f (x 0)

x −x 0 = limh →0

f (x 0+h )

−f (x 0)

h  .

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Differentiability and Derivative: If I  is an interval of R, then a

function f  : I → R is said to be differentiable at x 0 ∈ I  if

limx →x 0

f (x )−f (x 0)x −x 0

(or, equivalently limh →0

f (x 0+h )−f (x 0)h  ) exists in R.

If f  is differentiable at x 0, then the derivative of f  at x 0 is

f (x 

0) =limx →x 0

f (x )−f (x 0)

x −x 0 =limh →0

f (x 0+h )−f (x 0)

h .

f  : I → R is said to be differentiable if f  is differentiable at each

x 0 ∈ I .

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Differentiability and Derivative: If I  is an interval of R, then a

function f  : I → R is said to be differentiable at x 0 ∈ I  if

limx →x 0

f (x )−f (x 0)x −x 0

(or, equivalently limh →0

f (x 0+h )−f (x 0)h  ) exists in R.

If f  is differentiable at x 0, then the derivative of f  at x 0 is

f (x 

0) =limx →x 0

f (x )−f (x 0)

x −x 0 =limh →0

f (x 0+h )−f (x 0)

h .

f  : I → R is said to be differentiable if f  is differentiable at each

x 0 ∈ I .

Result: If f  : I 

→R is differentiable at x 0

∈I , then f  is

continuous at x 0.

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Examples:

1. For n = 1, 2, 3, let f n (x ) =

x n sin 1

x  if x = 0,

0 if x  = 0.

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Examples:

1. For n = 1, 2, 3, let f n (x ) =

x n sin 1

x  if x = 0,

0 if x  = 0.

2. f (x ) = x 2 if x ∈ Q,

0 if x ∈ R \Q.

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Examples:

1. For n = 1, 2, 3, let f n (x ) =

x n sin 1

x  if x = 0,

0 if x  = 0.

2. f (x ) = x 2 if x ∈ Q,

0 if x ∈ R \Q.

Ex. Let f (x ) = x 2|x | for all x ∈ R. Examine the existence of

f (x ), f (x ) and f (x ), where x ∈ R.

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Examples:

1. For n = 1, 2, 3, let f n (x ) =

x n sin 1

x  if x = 0,

0 if x  = 0.

2. f (x ) = x 2 if x ∈ Q,

0 if x ∈ R \Q.

Ex. Let f (x ) = x 2|x | for all x ∈ R. Examine the existence of

f (x ), f (x ) and f (x ), where x ∈ R.

Rules for finding derivatives:

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Definition: f  : D → R has a local maximum (resp. minimum) at

x 0

∈D  if there exists δ > 0 such that f (x )

≤f (x 0)

(resp. f (x 0) ≤ f (x )) for all x ∈ (x 0 − δ,x 0 + δ) ∩ D .

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Definition: f  : D → R has a local maximum (resp. minimum) at

x 0 ∈ D  if there exists δ > 0 such that f (x ) ≤ f (x 0)

(resp. f (x 0) ≤ f (x )) for all x ∈ (x 0 − δ,x 0 + δ) ∩ D .

Result: If f  : D → R has a local maximum or local minimum at

an interior point x 0 of D and if f  is differentiable at x 0, then

f (x 0) = 0.

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Definition: f  : D → R has a local maximum (resp. minimum) at

x 0 ∈ D  if there exists δ > 0 such that f (x ) ≤ f (x 0)

(resp. f (x 0) ≤ f (x )) for all x ∈ (x 0 − δ,x 0 + δ) ∩ D .

Result: If f  : D → R has a local maximum or local minimum at

an interior point x 0 of D and if f  is differentiable at x 0, then

f (x 0) = 0.

Rolle’s theorem: If f  : [a ,b ] → R is continuous, if f  isdifferentiable on (a ,b ) and if f (a ) = f (b ), then there exists

ξ ∈ (a ,b ) such that f (ξ) = 0.

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Definition: f  : D → R has a local maximum (resp. minimum) at

x 0 ∈ D  if there exists δ > 0 such that f (x ) ≤ f (x 0)

(resp. f (x 0) ≤ f (x )) for all x ∈ (x 0 − δ,x 0 + δ) ∩ D .

Result: If f  : D → R has a local maximum or local minimum at

an interior point x 0 of D and if f  is differentiable at x 0, then

f (x 0) = 0.

Rolle’s theorem: If f  : [a ,b ] → R is continuous, if f  isdifferentiable on (a ,b ) and if f (a ) = f (b ), then there exists

ξ ∈ (a ,b ) such that f (ξ) = 0.

Example: The equation x 2 = x sin x + cos x  has exactly two

real roots.

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Definition: f  : D → R has a local maximum (resp. minimum) at

x 0 ∈ D  if there exists δ > 0 such that f (x ) ≤ f (x 0)

(resp. f (x 0) ≤ f (x )) for all x ∈ (x 0 − δ,x 0 + δ) ∩ D .

Result: If f  : D → R has a local maximum or local minimum at

an interior point x 0 of D and if f  is differentiable at x 0, then

f (x 0) = 0.

Rolle’s theorem: If f  : [a ,b ] → R is continuous, if f  isdifferentiable on (a ,b ) and if f (a ) = f (b ), then there exists

ξ ∈ (a ,b ) such that f (ξ) = 0.

Example: The equation x 2 = x sin x + cos x  has exactly two

real roots.Ex. Find the number of real roots of the equation

x 4 + 2x 2 − 6x + 2 = 0.

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Mean value theorem: If f  : [a ,b ] → R is continuous and if f  is

differentiable on (a ,b ), then there exists ξ

∈(a ,b ) such that

f (b ) − f (a ) = f (ξ)(b − a ).

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Mean value theorem: If f  : [a ,b ] → R is continuous and if f  is

differentiable on (a ,b ), then there exists ξ

∈(a ,b ) such that

f (b ) − f (a ) = f (ξ)(b − a ).

Result: If f  : I → R is differentiable and f (x ) = 0 for all x ∈ I ,

then f  is constant on I .

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Mean value theorem: If f  : [a ,b ] → R is continuous and if f  is

differentiable on (a ,b ), then there exists ξ

∈(a ,b ) such that

f (b ) − f (a ) = f (ξ)(b − a ).

Result: If f  : I → R is differentiable and f (x ) = 0 for all x ∈ I ,

then f  is constant on I .

Result: Let f  : I →R be differentiable.

(i) If f (x ) ≥ 0 for all x ∈ I , then f  is increasing on I .

(ii) If f (x ) ≤ 0 for all x ∈ I , then f  is decreasing on I .

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Mean value theorem: If f  : [a ,b ] → R is continuous and if f  is

differentiable on (a ,b ), then there exists ξ

∈(a ,b ) such that

f (b ) − f (a ) = f (ξ)(b − a ).

Result: If f  : I → R is differentiable and f (x ) = 0 for all x ∈ I ,

then f  is constant on I .

Result: Let f  : I →R be differentiable.

(i) If f (x ) ≥ 0 for all x ∈ I , then f  is increasing on I .

(ii) If f (x ) ≤ 0 for all x ∈ I , then f  is decreasing on I .

Example: sin x ≥ x − x 3

6 for all x ∈ [0, π2 ].

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Mean value theorem: If f  : [a ,b ] → R is continuous and if f  is

differentiable on (a ,b ), then there exists ξ

∈(a ,b ) such that

f (b ) − f (a ) = f (ξ)(b − a ).

Result: If f  : I → R is differentiable and f (x ) = 0 for all x ∈ I ,

then f  is constant on I .

Result: Let f  : I →R be differentiable.

(i) If f (x ) ≥ 0 for all x ∈ I , then f  is increasing on I .

(ii) If f (x ) ≤ 0 for all x ∈ I , then f  is decreasing on I .

Example: sin x ≥ x − x 3

6 for all x ∈ [0, π2 ].

Ex. If f (x ) = x 3 + x 2 − 5x + 3 for all x ∈ R, then show that f  is

one-one on [1, 5] but not one-one on R.

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Intermediate value property of derivatives: Let f  : I → R be

differentiable and let a ,b ∈I with a < b . If f 

(a ) < k  < f 

(b ),

then there exists c ∈ (a ,b ) such that f (c ) = k .

Ex. Let f  : R→ R be differentiable such that f (−1) = 5,

f (0) = 0 and f (1) = 10. Prove that there exist ξ1, ξ2 ∈ (−1, 1)such that f (ξ1) =

−3 and f (ξ2) = 3.

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Intermediate value property of derivatives: Let f  : I → R be

differentiable and let a ,b ∈I with a < b . If f 

(a ) < k  < f 

(b ),

then there exists c ∈ (a ,b ) such that f (c ) = k .

Ex. Let f  : R→ R be differentiable such that f (−1) = 5,

f (0) = 0 and f (1) = 10. Prove that there exist ξ1, ξ2 ∈ (−1, 1)such that f (ξ1) =

−3 and f (ξ2) = 3.

Local maximum & Local minimum : Sufficient conditions:

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Intermediate value property of derivatives: Let f  : I → R be

differentiable and let a ,b ∈I with a < b . If f (a ) < k  < f (b ),

then there exists c ∈ (a ,b ) such that f (c ) = k .

Ex. Let f  : R→ R be differentiable such that f (−1) = 5,

f (0) = 0 and f (1) = 10. Prove that there exist ξ1, ξ2 ∈ (−1, 1)such that f (ξ1) =

−3 and f (ξ2) = 3.

Local maximum & Local minimum : Sufficient conditions:

1. First derivative test

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Intermediate value property of derivatives: Let f  : I → R be

differentiable and let a ,b ∈I with a < b . If f (a ) < k  < f (b ),

then there exists c ∈ (a ,b ) such that f (c ) = k .

Ex. Let f  : R→ R be differentiable such that f (−1) = 5,

f (0) = 0 and f (1) = 10. Prove that there exist ξ1, ξ2 ∈ (−1, 1)such that f (ξ1) =

−3 and f (ξ2) = 3.

Local maximum & Local minimum : Sufficient conditions:

1. First derivative test

2. Second derivative test

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Intermediate value property of derivatives: Let f  : I → R be

differentiable and let a ,b ∈I with a < b . If f (a ) < k  < f (b ),

then there exists c ∈ (a ,b ) such that f (c ) = k .

Ex. Let f  : R→ R be differentiable such that f (−1) = 5,

f (0) = 0 and f (1) = 10. Prove that there exist ξ1, ξ2 ∈ (−1, 1)such that f (ξ1) =

−3 and f (ξ2) = 3.

Local maximum & Local minimum : Sufficient conditions:

1. First derivative test

2. Second derivative test

Example: Find the local maxima and local minima of f , wheref (x ) = 1 − x 2/3 for all x ∈ R.

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L’Hopital’s rules:

1. Let f  : (a ,b ) → R and g  : (a ,b ) → R be differentiable at

x 0 ∈ (a ,b ). Also, let f (x 0) = g (x 0) = 0 and g (x 0) = 0. Thenlim

x →x 0

f (x )g (x ) = f (x 0)

g (x 0).

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L’Hopital’s rules:

1. Let f  : (a ,b ) → R and g  : (a ,b ) → R be differentiable at

x 0 ∈ (

a ,b ). Also, let

f (x 

0) =g 

(x 

0) = 0 andg (x 

0) = 0. Thenlimx →x 0

f (x )g (x ) = f (x 0)

g (x 0).

2. Let f  : (a ,b ) → R and g  : (a ,b ) → R be differentiable such

that limx →

a +f (x ) = lim

x →

a +g (x ) = 0 and g (x )

= 0 for all

x ∈ (a ,b ). If limx →a +

f (x )g (x ) = , then lim

x →a +

f (x )g (x ) = .

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L’Hopital’s rules:

1. Let f  : (a ,b ) → R and g  : (a ,b ) → R be differentiable at

x 0 ∈ (

a ,b ). Also, let f 

(x 

0) =g 

(x 

0) =0 and g 

(x 

0) =0. Then

limx →x 0

f (x )g (x ) = f (x 0)

g (x 0).

2. Let f  : (a ,b ) → R and g  : (a ,b ) → R be differentiable such

that limx →

a +f (x ) = lim

x →

a +g (x ) = 0 and g (x )

= 0 for all

x ∈ (a ,b ). If limx →a +

f (x )g (x ) = , then lim

x →a +

f (x )g (x ) = .

Examples: (i) limx →0

√1+x −1x  (ii) lim

x →π2

1−sin x 1+cos2x 

(iii) limx →0

x 2

sin1

x sin x  (iv) lim

x →0( sin x 

x  ) 1x  (v) limx →0

e −

1/x 2

(vi) limx →0

( 1sin x  − 1

x ) (vii) limx →∞

x −sin x 2x +sin x 

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Taylor’s theorem: Let f  : [a ,b ]→R be such that f , f , f , ..., f (n )

are continuous on [a ,b ] and f (n +1) exists on (a ,b ).

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Taylor’s theorem: Let f  : [a ,b ]→R be such that f , f , f , ..., f (n )

are continuous on [a ,b ] and f (n +1) exists on (a ,b ). Then there

exists ξ ∈ (a ,b ) such that

f (b ) = f (a ) + f (a )(b − a ) + f (a )2! (b − a )2 + · · · + f (n )(a )

n ! (b − a )n 

+

f (n +1)(ξ)

(n +1)! (b −

a )

n +1

.

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Taylor’s theorem: Let f  : [a ,b ]→R be such that f , f , f , ..., f (n )

are continuous on [a ,b ] and f (n +1) exists on (a ,b ). Then there

exists ξ ∈ (a ,b ) such that

f (b ) = f (a ) + f (a )(b − a ) + f (a )2! (b − a )2 + · · · + f (n )(a )

n ! (b − a )n 

+

f (n +1)(ξ)

(n +1)! (b −

a )

n +1.

Example: 1 + x 2 − x 2

8 ≤ √1 + x ≤ 1 + x 

2 for all x  > 0.

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Taylor’s theorem: Let f  : [a ,b ]

→R be such that f , f , f , ..., f (n )

are continuous on [a ,b ] and f (n +1) exists on (a ,b ). Then there

exists ξ ∈ (a ,b ) such that

f (b ) = f (a ) + f (a )(b − a ) + f (a )2! (b − a )2 + · · · + f (n )(a )

n ! (b − a )n 

+

f (n +1)(ξ)

(n +1)! (b −

a )

n +1.

Example: 1 + x 2 − x 2

8 ≤ √1 + x ≤ 1 + x 

2 for all x  > 0.

Taylor series & Maclaurin series: Convergence

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Taylor’s theorem: Let f  : [a ,b ]

→R be such that f , f , f , ..., f (n )

are continuous on [a ,b ] and f (n +1) exists on (a ,b ). Then there

exists ξ ∈ (a ,b ) such that

f (b ) = f (a ) + f (a )(b − a ) + f (a )2! (b − a )2 + · · · + f (n )(a )

n ! (b − a )n 

+ f (n +1)(ξ)

(n +1)!(b −

a )n +1.

Example: 1 + x 2 − x 2

8 ≤ √1 + x ≤ 1 + x 

2 for all x  > 0.

Taylor series & Maclaurin series: Convergence

Example: Taylor series expansion of e x , sin x  and cos x .

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Result on local maxima and local minima:

Let x 0∈

(a ,b ) and let n 

≥2. Also, let f , f , ..., f (n ) be continuous

on (a ,b ) and f (x 0) = f (x 0) = · · · = f (n −1)(x 0) = 0 butf (n )(x 0) = 0.

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Result on local maxima and local minima:

Let x 0∈

(a ,b ) and let n 

≥2. Also, let f , f , ..., f (n ) be continuous

on (a ,b ) and f (x 0) = f (x 0) = · · · = f (n −1)(x 0) = 0 butf (n )(x 0) = 0.

(i) If n  is even and f (n )(x 0) < 0, then f  has a local maximum

at x 0.

(ii) If n  is even and f (n )(x 0) > 0, then f  has a local minimum atx 0.

(iii) If n  is odd, then f  has neither a local maximum nor a local

minimum at x 0.

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Result on local maxima and local minima:

Let x 0∈

(a ,b ) and let n 

≥2. Also, let f , f , ..., f (n ) be continuous

on (a ,b ) and f (x 0) = f (x 0) = · · · = f (n −1)(x 0) = 0 butf (n )(x 0) = 0.

(i) If n  is even and f (n )(x 0) < 0, then f  has a local maximum

at x 0.

(ii) If n  is even and f (n )(x 0) > 0, then f  has a local minimum atx 0.

(iii) If n  is odd, then f  has neither a local maximum nor a local

minimum at x 0.

Ex. Find all the local maximum and local minimum values of f ,where f (x ) = x 5 − 5x 4 + 5x 3 + 12 for all x ∈ R.