SLecture3

MA 105 Calculus: Lecture 3Differentiation

S. Sivaji Ganesh

Mathematics DepartmentIIT Bombay

July 28, 2009

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 1 / 24

Derivatives

Definition

The derivative of a function f at a,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation :

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f ,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f , draw the line joining the points(a, f (a)), (x , f (x)).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f , draw the line joining the points(a, f (a)), (x , f (x)). Take its slope

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f , draw the line joining the points(a, f (a)), (x , f (x)). Take its slope and take the limit of these slopes asx → a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f , draw the line joining the points(a, f (a)), (x , f (x)). Take its slope and take the limit of these slopes asx → a. Then the point (x , f (x)) tends to (a, f (a)).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f , draw the line joining the points(a, f (a)), (x , f (x)). Take its slope and take the limit of these slopes asx → a. Then the point (x , f (x)) tends to (a, f (a)). the limit is nothingbut

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f , draw the line joining the points(a, f (a)), (x , f (x)). Take its slope and take the limit of these slopes asx → a. Then the point (x , f (x)) tends to (a, f (a)). the limit is nothingbut the slope of the tangent line at (a, f (a))

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Definition

The derivative of a function f at a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists. We say f is differentiable at a.

If we write x = a + h, then h = x − a and h → 0 if and only if x → a.Thus

f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f , draw the line joining the points(a, f (a)), (x , f (x)). Take its slope and take the limit of these slopes asx → a. Then the point (x , f (x)) tends to (a, f (a)). the limit is nothingbut the slope of the tangent line at (a, f (a)) to the curve y = f (x).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 2 / 24

Derivatives

Example

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 3 / 24

Derivatives

Example Considerf (x) = x3 − 2

.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 3 / 24

Derivatives

Example Considerf (x) = x3 − 2

.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 3 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate, and Guess!

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate, and Guess!

1 f (x) = x2

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate, and Guess!

1 f (x) = x2

2 f (x) = |x |

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate, and Guess!

1 f (x) = x2

2 f (x) = |x |3 f (x) = |x |2

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate, and Guess!

1 f (x) = x2

2 f (x) = |x |3 f (x) = |x |2

4 f (x) = x3

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate, and Guess!

1 f (x) = x2

2 f (x) = |x |3 f (x) = |x |2

4 f (x) = x3

5 f (x) = sin x

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivatives

DefinitionA function f is said to be differentiable on (a,b) if f is differentiable atevery point in (a,b).

What is the derivative at x = 0 of the following functions? Look at theirgraphs, use the interpretation given for derivate, and Guess!

1 f (x) = x2

2 f (x) = |x |3 f (x) = |x |2

4 f (x) = x3

5 f (x) = sin x

Now, Use the definition of f ′(0) to compute.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 4 / 24

Derivative as a function

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

The domain of f ′ may be smaller than the domain of f .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

The domain of f ′ may be smaller than the domain of f .Notation

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

The domain of f ′ may be smaller than the domain of f .Notation Some of the common notations are

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

The domain of f ′ may be smaller than the domain of f .Notation Some of the common notations are f ′(x),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

The domain of f ′ may be smaller than the domain of f .Notation Some of the common notations are f ′(x), df

dx (x),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

The domain of f ′ may be smaller than the domain of f .Notation Some of the common notations are f ′(x), df

dx (x), dydx

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Derivative as a function

Let f be a function. collect all a at which f is differentiable i.e., the setS := {a : f ′(a) exists}. On this set, we can define a new function. Thefunction which assigns to each x ∈ S, the number f ′(x).

f ′(x) = limh→0

f (x + h) − f (x)

h

The domain of f ′ may be smaller than the domain of f .Notation Some of the common notations are f ′(x), df

dx (x), dydx (y = f (x)

indicates that x is an independent variable and y is the dependentvariable.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 5 / 24

Example (contd.)

Example

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous? Where?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous? Where? How to prove?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous? Where? How to prove? What are all the numbers atwhich f ′ exists?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous? Where? How to prove? What are all the numbers atwhich f ′ exists? How to prove?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous? Where? How to prove? What are all the numbers atwhich f ′ exists? How to prove?

Now justify your answers

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous? Where? How to prove? What are all the numbers atwhich f ′ exists? How to prove?

Now justify your answers for x = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 6 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function, we can guess the sign of thederivative function at every number.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function, we can guess the sign of thederivative function at every number. The following theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function, we can guess the sign of thederivative function at every number. The following theorem withg(x) = 0 may be useful?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function, we can guess the sign of thederivative function at every number. The following theorem withg(x) = 0 may be useful?

TheoremIf f (x) ≤ (≥)g(x) when x is near a (except possibly at a) and the limitsof f and g both exist as x approaches a, then

limx→a

f (x) ≤ (≥) limx→a

g(x).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function, we can guess the sign of thederivative function at every number. The following theorem withg(x) = 0 may be useful?

TheoremIf f (x) ≤ (≥)g(x) when x is near a (except possibly at a) and the limitsof f and g both exist as x approaches a, then

limx→a

f (x) ≤ (≥) limx→a

g(x).

Suppose we are given graph of a function φ,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function, we can guess the sign of thederivative function at every number. The following theorem withg(x) = 0 may be useful?

TheoremIf f (x) ≤ (≥)g(x) when x is near a (except possibly at a) and the limitsof f and g both exist as x approaches a, then

limx→a

f (x) ≤ (≥) limx→a

g(x).

Suppose we are given graph of a function φ, what should be my choicefor f in the above theorem?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

By looking at the graph of the function, we can guess the sign of thederivative function at every number. The following theorem withg(x) = 0 may be useful?

TheoremIf f (x) ≤ (≥)g(x) when x is near a (except possibly at a) and the limitsof f and g both exist as x approaches a, then

limx→a

f (x) ≤ (≥) limx→a

g(x).

Suppose we are given graph of a function φ, what should be my choicefor f in the above theorem? Newton quotients .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 7 / 24

Can we guess the values/Graph of derivative function?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 8 / 24

Can we guess the values/Graph of derivative function?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 9 / 24

Can we guess the values/Graph of derivative function?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 10 / 24

Derivatives

TheoremIf f is differentiable at a,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a. Q.E.D.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a. Q.E.D.Is the converse true?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a. Q.E.D.Is the converse true? The converse is not true.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a. Q.E.D.Is the converse true? The converse is not true. Give an example.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a. Q.E.D.Is the converse true? The converse is not true. Give an example. Giveanother example,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a. Q.E.D.Is the converse true? The converse is not true. Give an example. Giveanother example, it should not be

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Derivatives

TheoremIf f is differentiable at a, then f is continuous at a.

Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)

f (x) =f (x) − f (a)

x − a(x − a) + f (a)

What is next? Take limit as x → a. Q.E.D.Is the converse true? The converse is not true. Give an example. Giveanother example, it should not be f (x) = |x |.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 11 / 24

Reasons for non-differentiability of functions at a

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

There are three ways

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

There are three ways in which a function fails to be differentiable at a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

There are three ways in which a function fails to be differentiable at a.1 Function is not continuous at a

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

There are three ways in which a function fails to be differentiable at a.1 Function is not continuous at a (by previous theorem)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

There are three ways in which a function fails to be differentiable at a.1 Function is not continuous at a (by previous theorem)2 There is corner (in the graph) at (a, f (a)).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

There are three ways in which a function fails to be differentiable at a.1 Function is not continuous at a (by previous theorem)2 There is corner (in the graph) at (a, f (a)). (as in |x | case)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

There are three ways in which a function fails to be differentiable at a.1 Function is not continuous at a (by previous theorem)2 There is corner (in the graph) at (a, f (a)). (as in |x | case)3 Tangent at (a, f (a) to the graph is vertical. (next slide)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 12 / 24

Reasons for non-differentiability of functions at a

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 13 / 24

Reasons for non-differentiability of functions at a

f (x) = x1/3 with domain R

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 13 / 24

Reasons for non-differentiability of functions at a

f (x) = x1/3 with domain R

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 13 / 24

Reasons for non-differentiability of functions at a

f (x) = x1/3 with domain R

vertical tangent at x = 0S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 13 / 24

Reasons for non-differentiability of functions at a

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 14 / 24

Reasons for non-differentiability of functions at a

f (x) = x2/3 with domain R

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 14 / 24

Reasons for non-differentiability of functions at a

f (x) = x2/3 with domain R

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 14 / 24

Reasons for non-differentiability of functions at a

f (x) = x2/3 with domain R

vertical tangent at x = 0S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 14 / 24

Derivatives of Polynomials and Exponential functions

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good!

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number. Find out the definition of ax ,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number. Find out the definition of ax ,after thinking for a while what it should be!

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number. Find out the definition of ax ,after thinking for a while what it should be!

DefinitionThe number e is that real number

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number. Find out the definition of ax ,after thinking for a while what it should be!

DefinitionThe number e is that real number for which

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number. Find out the definition of ax ,after thinking for a while what it should be!

DefinitionThe number e is that real number for which the graph of the function ex

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number. Find out the definition of ax ,after thinking for a while what it should be!

DefinitionThe number e is that real number for which the graph of the function ex

has slope 1

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Derivatives of Polynomials and Exponential functions

For n ∈ N, prove the formula

ddx

xn = nxn−1

using the definition. What if n ∈ R? State and prove.Exponential functions Let a > 0, let f (x) = ax . Then

f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good! what is the meaning of ax when x is a real number? I know if xis a natural number or a rational number. Find out the definition of ax ,after thinking for a while what it should be!

DefinitionThe number e is that real number for which the graph of the function ex

has slope 1 at x = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 15 / 24

Differentiation Rules

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 16 / 24

Differentiation Rules

TheoremLet f and g be differentiable at a. Then f + g, fg are differentiable at a,and so is f/g if g(a) 6= 0, and

1 (f + g)′(a) = f ′(a) + g′(a)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 16 / 24

Differentiation Rules

TheoremLet f and g be differentiable at a. Then f + g, fg are differentiable at a,and so is f/g if g(a) 6= 0, and

1 (f + g)′(a) = f ′(a) + g′(a)

2 (fg)′(a) = f ′(a)g(a) + f (a)g′(a)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 16 / 24

Differentiation Rules

TheoremLet f and g be differentiable at a. Then f + g, fg are differentiable at a,and so is f/g if g(a) 6= 0, and

1 (f + g)′(a) = f ′(a) + g′(a)

2 (fg)′(a) = f ′(a)g(a) + f (a)g′(a)

3 ( fg )′(a) = g(a)f ′(a)−f (a)g′(a)

(g(a))2

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 16 / 24

Differentiation Rules

TheoremLet f and g be differentiable at a. Then f + g, fg are differentiable at a,and so is f/g if g(a) 6= 0, and

1 (f + g)′(a) = f ′(a) + g′(a)

2 (fg)′(a) = f ′(a)g(a) + f (a)g′(a)

3 ( fg )′(a) = g(a)f ′(a)−f (a)g′(a)

(g(a))2 if g(a) 6= 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 16 / 24

Differentiation Rules

TheoremLet f and g be differentiable at a. Then f + g, fg are differentiable at a,and so is f/g if g(a) 6= 0, and

1 (f + g)′(a) = f ′(a) + g′(a)

2 (fg)′(a) = f ′(a)g(a) + f (a)g′(a)

3 ( fg )′(a) = g(a)f ′(a)−f (a)g′(a)

(g(a))2 if g(a) 6= 0.

Proof: Note

f (x)g(x) − f (a)g(a) = f (x)[g(x) − g(a)] + g(a)[f (x) − f (a)]

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 16 / 24

Differentiation Rules

TheoremLet f and g be differentiable at a. Then f + g, fg are differentiable at a,and so is f/g if g(a) 6= 0, and

1 (f + g)′(a) = f ′(a) + g′(a)

2 (fg)′(a) = f ′(a)g(a) + f (a)g′(a)

3 ( fg )′(a) = g(a)f ′(a)−f (a)g′(a)

(g(a))2 if g(a) 6= 0.

Proof: Note

f (x)g(x) − f (a)g(a) = f (x)[g(x) − g(a)] + g(a)[f (x) − f (a)]

f (x)

g(x)−

f (a)

g(a)=

1g(x)g(a)

[g(a) (f (x) − f (a)) − f (a) (g(x) − g(a))]

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 16 / 24

Derivatives

TheoremSuppose f is differentiable at a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 17 / 24

Derivatives

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 17 / 24

Derivatives

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 17 / 24

Derivatives

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.

Proof: Define φ by

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 17 / 24

Derivatives

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.

Proof: Define φ by

φ(x) =f (x) − f (a)

x − a− f ′(a).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 17 / 24

Derivatives

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.

Proof: Define φ by

φ(x) =f (x) − f (a)

x − a− f ′(a).

Since f is differentiable at a,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 17 / 24

Derivatives

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.

Proof: Define φ by

φ(x) =f (x) − f (a)

x − a− f ′(a).

Since f is differentiable at a, the result follows.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 17 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) =

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.Once again, by previous theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.Once again, by previous theorem applied to f at the point g(a),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.Once again, by previous theorem applied to f at the point g(a), wehave

f (t) − f (g(a)) = (t − g(a))f ′(g(a)) + (t − g(a))ψ(t),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.Once again, by previous theorem applied to f at the point g(a), wehave

f (t) − f (g(a)) = (t − g(a))f ′(g(a)) + (t − g(a))ψ(t),

with limt→g(a) ψ(t) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.Once again, by previous theorem applied to f at the point g(a), wehave

f (t) − f (g(a)) = (t − g(a))f ′(g(a)) + (t − g(a))ψ(t),

with limt→g(a) ψ(t) = 0.Now

f (g(x)) − f (g(a)) = [g(x) − g(a)].[f ′(g(a)) + ψ(g(x))]

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.Once again, by previous theorem applied to f at the point g(a), wehave

f (t) − f (g(a)) = (t − g(a))f ′(g(a)) + (t − g(a))ψ(t),

with limt→g(a) ψ(t) = 0.Now

f (g(x)) − f (g(a)) = [g(x) − g(a)].[f ′(g(a)) + ψ(g(x))]

f (g(x)) − f (g(a)) = (x − a)g′(a) + (x − a)φ(x).[f ′(g(a)) + ψ(g(x))]S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a and f is differentiable at g(a). Then f◦gis differentiable at a and (f◦g)′(a) = f ′(g(a))g′(a).

Proof: By previous theorem applied to g, we have

g(x) − g(a) = (x − a)g′(a) + (x − a)φ(x),

with limx→a φ(x) = 0.Once again, by previous theorem applied to f at the point g(a), wehave

f (t) − f (g(a)) = (t − g(a))f ′(g(a)) + (t − g(a))ψ(t),

with limt→g(a) ψ(t) = 0.Now

f (g(x)) − f (g(a)) = [g(x) − g(a)].[f ′(g(a)) + ψ(g(x))]

f (g(x)) − f (g(a)) = (x − a)g′(a) + (x − a)φ(x).[f ′(g(a)) + ψ(g(x))]S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 18 / 24

Chain Rule (contd.)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 19 / 24

Chain Rule (contd.)

f (g(x)) − f (g(a)) = (x − a)g′(a) + (x − a)φ(x).[f ′(g(a)) + ψ(g(x))]

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 19 / 24

Chain Rule (contd.)

f (g(x)) − f (g(a)) = (x − a)g′(a) + (x − a)φ(x).[f ′(g(a)) + ψ(g(x))]

Now take limit as x → a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 19 / 24

Problems

Consider

f (x) =

{

x sin(1x ) if x 6= 0

0 if x = 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 20 / 24

Problems

Consider

f (x) =

{

x sin(1x ) if x 6= 0

0 if x = 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 20 / 24

Problems

It is differentiable for x 6= 0 by Chain rule.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

How about the function

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

How about the function

g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

How about the function

g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

Is f differentiable?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

How about the function

g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

Is f differentiable? Is f ′ continuous?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

How about the function

g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

Is f differentiable? Is f ′ continuous? Does f ′ has the intermediate valueproperty?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

How about the function

g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

Is f differentiable? Is f ′ continuous? Does f ′ has the intermediate valueproperty?

Fact: If f is differentiable in an interval,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Problems

It is differentiable for x 6= 0 by Chain rule. How about at x = 0? notdifferentiable at x = 0. Prove.

How about the function

g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

Is f differentiable? Is f ′ continuous? Does f ′ has the intermediate valueproperty?

Fact: If f is differentiable in an interval, then f ′ has the intermediatevalue property.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 21 / 24

Implicit differentiation

We know how to do this.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 22 / 24

Implicit differentiation

We know how to do this. Proof will be seen when we do

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 22 / 24

Implicit differentiation

We know how to do this. Proof will be seen when we do multivariableChain rule!

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 22 / 24

Derivative of Inverse function

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that the inverse function f−1 is also differentiable on its domain?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that the inverse function f−1 is also differentiable on its domain?Answer is No.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that the inverse function f−1 is also differentiable on its domain?Answer is No.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that the inverse function f−1 is also differentiable on its domain?Answer is No.

there is a vertical tangent at x = 0 for the function x1/3.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that the inverse function f−1 is also differentiable on its domain?Answer is No.

there is a vertical tangent at x = 0 for the function x1/3.What do we learn from the above example?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that the inverse function f−1 is also differentiable on its domain?Answer is No.

there is a vertical tangent at x = 0 for the function x1/3.What do we learn from the above example? Can we apply chain ruleto f−1◦f (x) = x?S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function

Question: If f is an invertible function, and f is differentiable , then is ittrue that the inverse function f−1 is also differentiable on its domain?Answer is No.

there is a vertical tangent at x = 0 for the function x1/3.What do we learn from the above example? Can we apply chain ruleto f−1◦f (x) = x? Explain your answer.S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 23 / 24

Derivative of Inverse function (contd.)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 24 / 24

Derivative of Inverse function (contd.)

there is a vertical tangent at x = 0 for the function x1/3.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 24 / 24

Derivative of Inverse function (contd.)

there is a vertical tangent at x = 0 for the function x1/3.

What do we learn from the above example?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 24 / 24

Derivative of Inverse function (contd.)

there is a vertical tangent at x = 0 for the function x1/3.

What do we learn from the above example?Can we apply chain rule to

f−1◦f (x) = x?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 24 / 24

Derivative of Inverse function (contd.)

there is a vertical tangent at x = 0 for the function x1/3.

What do we learn from the above example?Can we apply chain rule to

f−1◦f (x) = x?

Explain your answer.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 3 July 28, 2009 24 / 24