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,


Derivatives

Definition

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


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h

if this limit exists.


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h

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


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h


If we write x = a + h,


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h


If we write x = a + h, then h = x − a


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h


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


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h


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


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h


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


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h


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.


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation :


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



f ′(a) = limx→a

f (x) − f (a)

x − a.

Interpretation : Take the graph of f ,


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



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)).


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



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


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



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.


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



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)).


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



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


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



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))


Derivatives

Definition


f ′(a) = limh→0

f (a + h) − f (a)

h



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).


Derivatives

Example


Derivatives

Example Considerf (x) = x3 − 2

.


Derivatives

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


Derivatives


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


Derivatives


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


Derivatives


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


Derivatives


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


Derivatives



1 f (x) = x2


Derivatives



1 f (x) = x2

2 f (x) = |x |


Derivatives



1 f (x) = x2

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


Derivatives



1 f (x) = x2

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

4 f (x) = x3


Derivatives



1 f (x) = x2

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

4 f (x) = x3

5 f (x) = sin x


Derivatives



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.


Derivative as a function



Let f be a function.



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



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



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.



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,



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




f ′(x) = limh→0

f (x + h) − f (x)

h

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




f ′(x) = limh→0

f (x + h) − f (x)

h

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




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) = 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),




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),




f ′(x) = limh→0

f (x + h) − f (x)

h


dx (x), dydx




f ′(x) = limh→0

f (x + h) − f (x)

h


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

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


Example (contd.)

Example


Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

.


Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8

. Is f continuous?


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?


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?


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?


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?


Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8


Now justify your answers


Example (contd.)

Example Consider

f (x) =

−x if x ≤ 0

x if 0 < x < 4

−x + 8 if 4 < x < 8


Now justify your answers for x = 0.


Can we guess the values/Graph of derivative function?

By looking at the graph of the function,



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



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



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).





limx→a

f (x) ≤ (≥) limx→a

g(x).

Suppose we are given graph of a function φ,





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?





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 .


Derivatives

TheoremIf f is differentiable at a,


Derivatives

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


Derivatives


Proof:

f (x) − f (a) =f (x) − f (a)

x − a(x − a)


Derivatives


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)


Derivatives


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?


Derivatives


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.


Derivatives


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.


Derivatives


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?


Derivatives


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.


Derivatives


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.


Derivatives


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,


Derivatives


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


Derivatives


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 |.


Reasons for non-differentiability of functions at a



There are three ways



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



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



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)).



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)



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)



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



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



For n ∈ N, prove the formula

ddx

xn = nxn−1




ddx

xn = nxn−1

using the definition.




ddx

xn = nxn−1

using the definition. What if n ∈ R?




ddx

xn = nxn−1

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




ddx

xn = nxn−1

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




ddx

xn = nxn−1

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




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




ddx

xn = nxn−1


f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax

Good!




ddx

xn = nxn−1


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




ddx

xn = nxn−1


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?




ddx

xn = nxn−1


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




ddx

xn = nxn−1


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.




ddx

xn = nxn−1


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 ,




ddx

xn = nxn−1


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!




ddx

xn = nxn−1


f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax


DefinitionThe number e is that real number




ddx

xn = nxn−1


f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax


DefinitionThe number e is that real number for which




ddx

xn = nxn−1


f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax


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




ddx

xn = nxn−1


f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax



has slope 1




ddx

xn = nxn−1


f ′(x) = limh→0

f (x + h) − f (x)

h= lim

h→0

ax (ah − 1)

h= f ′(0)ax



has slope 1 at x = 0.


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)




1 (f + g)′(a) = f ′(a) + g′(a)

2 (fg)′(a) = f ′(a)g(a) + f (a)g′(a)




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




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.




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)]




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))]


Derivatives

TheoremSuppose f is differentiable at a.


Derivatives

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


Derivatives


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


Derivatives


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

andlimx→a

φ(x) = 0.

Proof: Define φ by


Derivatives


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).


Derivatives


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,


Derivatives


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.


Chain Rule

Theorem (Chain Rule)

Suppose g is differentiable at a


Chain Rule


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


Chain Rule


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


Chain Rule


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


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)


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) =


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).


Chain Rule



Proof:


Chain Rule



Proof: By previous theorem


Chain Rule



Proof: By previous theorem applied to g,


Chain Rule



Proof: By previous theorem applied to g, we have

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


Chain Rule




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

with limx→a φ(x) = 0.


Chain Rule




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

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


Chain Rule




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),


Chain Rule




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),


Chain Rule




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



with limt→g(a) ψ(t) = 0.


Chain Rule




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



with limt→g(a) ψ(t) = 0.Now

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


Chain Rule




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



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.)


Chain Rule (contd.)

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


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.


Problems

Consider

f (x) =

{

x sin(1x ) if x 6= 0

0 if x = 0


Problems

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


Problems

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


Problems

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


Problems

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


Problems


How about the function


Problems



g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0


Problems



g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

Is f differentiable?


Problems



g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0

Is f differentiable? Is f ′ continuous?


Problems



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?


Problems



g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0


Fact: If f is differentiable in an interval,


Problems



g(x) =

{

x2 sin(1x ) if x 6= 0

0 if x = 0


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


Implicit differentiation

We know how to do this.



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



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


Derivative of Inverse function



Question: If f is an invertible function,



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



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



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?



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.




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




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



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.)




What do we learn from the above example?




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

f−1◦f (x) = x?




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

f−1◦f (x) = x?

Explain your answer.


SLecture3

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