Calculus on Rn

Seminar 7 2020

Differential Calculus. Part 2

1 The Differential for Branch Functions

This section contains examples of computing the differential function attached to real
functions of vector variable, so the codomain is R. Unlike the exercises proposed for
study in Seminar 6, the functions considered now are defined on branches. On the interior
of each branch, the study goes smooth, but at the meeting point of the branches, the
differentiability might suffer. In examples it is easy not notice all possible cases.
This is actually the place where we have to apply the entire algorithm for analyzing
the differentiability of a function at point. Let us once more recall the Algorithm:

Algorithm fornReal Functions of Vector Variables

 A ⊆ Rn


a ∈ intA
f :A→R


Step 1: Study if f has partial derivatives at a with respect to all variables.


Y ES → go to Step 2
N O → STOP the function does not have a differential at a
In fact, we have to determine
∂f
(a) ∀j ∈ {1, ..., n}.
∂xj

Step 2: Study the limit

 
1
l = lim f (a + h) − f (a) − hh, ∇f (a)i . (1)
h→0n khk

Recall that a = (a1 , ..., an ) and h = (h1 , ..., hn ) ∈ Rn

q
a + h = (a1 + h1 , ..., an + hn ) and khk = h21 + ... + h2n .
Moreover, the gradient of a function at a point is the vector composed of all of its partial
derivatives at that point. Thus
n
∂f ∂f X ∂f
hh, ∇f (a)i = h1 · (a) + ... + hn · (a) = hj · (a).
∂x1 ∂xn j=1
∂x j

1
This means that the limit (1) is actually
Pn ∂f
f (a1 + h1 , ..., a2 + h2 ) − f (a1 , ..., an ) − j=1 hj · ∂xj
(a)
l = lim p . (2)
h→0n h21 + ... + h2n

We encounter the following cases


l=0 → go to Step 3
otherwise → STOP the function does not have a differential at a

Step 3: We are in the case when l = 0. This means that the function f is differentiable
at the point a. Its differential is the linear function

df (a) : Rn → R

defined by
n
X ∂f
df (a)(h) = hh, ∇f (a)i = hj · (a), ∀h ∈ Rn .
j=1
∂xj

Example 1.1: Study the differentiability of the function

( 3 3
x −y
2 2 +y 2 (x, y) 6= (0, 0)
f : R → R, f (x, y) = x
0 (x, y) = (0, 0)

Solution:
We notice that the obvious problem appears at the point (0, 0). Let us consider the set

A := R2 \{(0, 0)}

This set is open. We may compute the partial derivatives of f with respect to both x and
y at each random point. Let (x, y) ∈ A be randomly chosen. Then

∂f x4 + 3x2 y 2 + 2xy 3
(x, y) =
∂x (x2 + y 2 )2

and
∂f y 4 + 3x2 y 2 + 2x3 y
(x, y) = − .
∂y (x2 + y 2 )2
These functions are continuous, therefore, f is differentiable on A.
The following step is to study the existence of the partial derivatives at (0, 0), starting
with respect to x:
x3 −0
∂f f (x, 0) − f (0, 0) 2
(0, 0) = lim = lim x +0 = 1 ∈ R .
∂x x→0 x−0 x→0 x − 0

This means that f has a partial derivative at (0, 0) with respect to x.

2
 Then we study the existence of the partial derivatives at (0, 0) with respect to y:
0−y 3
∂f f (0, y) − f (0, 0) 0+y 2 −y 3
(0, 0) = lim = lim = 3 = −1 ∈ R .
∂y x→0 x−0 y→0 y − 0 y
This means that f has a partial derivative at (0, 0) with respect to y.
We follow now the steps of the Algorithm from Section 4 of Seminar 6. We
are currently at Step 2. Recall that if the function were is differentiable at (0, 0),
the differential would be the function

df (0, 0) : R2 → R,

defined by
∂f ∂f
df (0, 0)(h1 , h2 ) = h(h1 , h2 , ), ∇f (0, 0)i = h1 (0, 0) + h2 (0, 0), ∀(h1 , h2 ) ∈ R2 .
∂x ∂y

WE DO NOT KNOW IF f IS DIFFERENTIABLE AT (0, 0), SO

In order to study the differentiability of f at (0, 0) we have to study the limit

   
∂f ∂f
f (0, 0) + (h1 , h2 ) − f (0, 0) − h1 ∂x (0, 0) + h2 ∂y (0, 0)
l= lim .
(h1 ,h2 )→(0,0) k(h1 , h2 ) − (0, 0)k
In order for the things to be simpler to analyse, we denote by

ω(h1 , h2 )

the value of a function at a point (h1 , h2 ) ∈ R2 generated by the limit that we have to
compute. So

We show that
6∃ lim ω(h1 , h2 ),
(h1 ,h2 )→(0,0)

with the help of sequences. So we consider the sequences, having as general terms, for all
k∈N    
1 1 2 1
ak = , and bk = , .
k k k k
We have that
lim ak = lim bk = (0, 0),
k→∞ k→∞

3
but

so
lim ω(ak ) 6= lim ω(bk ).
k→∞ k→∞

Therefore
6∃ lim ω(h1 , h2 )
(h1 ,h2 )→(0,0)

so the limit that we are looking for does not exist. This means that the function f
is not differentiable at (0, 0).
In conclusion the function f is differentiable on R2 \{(0, 0)}.

Example 1.2: Study the differentiability of the function

 4
x 3 sin xy x =
6 0
f : R2 → R, f (x, y) =
0 x=0

Solution:

To begin with, we see that the problem is this time represented by a large set..., not
just the point (0, 0), like it was the case in the other exercise. Set now

B := {(x, y) ∈ R2 : x 6= 0}.

Is is an open set, on which f has partial derivatives with respect to both x and y at each
given point (x, y) randomly chosen.

and

These two partial derivatives are continuous functions on B, this means that the
function f is differentiable on B. (We are not specifically asked to write the expression
of the differential function).
Part 2 We are left to study the differentiablity of f on R \B. Consider a random
point in this set,
(0, a) with a ∈ R .
∂f
First we analyze the partial derivatives. For ∂x
(0, a) we have to analyze

4
 ∂f
while for ∂y
(0, a) we have to analyze

We arrive at the conlusion that both partial derivatives at (0, a) exist, and

We have arrived at the Step 2 of the Algorithm, in which we analyze

Once again we make use of the notation ω the function whose limit we have to compute
now at (0, a), thus we are interested in

with

In this case we will prove that f is indeed differentiable at (0, a), so the limit
has to be 0. We make use of the sandwich theorem. We distinguish two cases:
Case 1: h1 6= 0, when

5
Case 2: h1 = 0, when f (0, a + h2 ) = 0.
From the Cases 1 and 2, we conclude that
1
kω(h1 , h2 )k ≤ kh1 k 3 , ∀(h1 , h2 ) ∈ R2

Therefore,
1
lim kω(h1 , h2 )k ≤ lim kh1 k 3 = 0
(h1 ,h2 )→(0,0) (h1 ,h2 )→(0,0)

So, the limit l of the Step 2 of the Algorithm exists, and is equal to 0. This means that
the function f is differentiable at (0, a) and the differential is the fucntion

df (0, a) : R2 → R

defined by

df (0, a)(h1 , h2 ) = h(h1 , h2 ), ∇f (0, a)i = h1 · 0 + h2 · 0 = 0, ∀(h1 , h2 ) ∈ R2 .

In conclusion, f is differentiable on R2 .

Example 1.3: Study the differentiability of the function

(
xy(x2 −y 2
2 x2 +y 2 (x, y) 6= (0, 0)
f : R → R, f (x, y) =
0 (x, y) = (0, 0)

Solution:
We notice that the obvious problem appears at the point (0, 0). Let us consider the set

A := R2 \{(0, 0)}

This set is open. We may compute the partial derivatives of f with respect to both x and
y at each random point. Let (x, y) ∈ A be randomly chosen. Then

∂f x4 y + 4x2 y 3 − y 5
(x, y) =
∂x (x2 + y 2 )2

and
∂f x5 − 4x3 y 2 − xy 4
(x, y) = .
∂y (x2 + y 2 )2
These functions are continuous, therefore, f is differentiable on A.
The following step is to study the existence of the partial derivatives at (0, 0), starting
with respect to x:
x·0(x−0)−0
∂f f (x, 0) − f (0, 0) x2 +0 0
(0, 0) = lim = lim = lim = lim 0 = 0 ∈ R .
∂x x→0 x−0 x→0 x−0 x→0 x3 x→0

This means that f has a partial derivative at (0, 0) with respect to x.

6
 Then we study the existence of the partial derivatives at (0, 0) with respect to y:
0·y(0−y 3 )
∂f f (0, y) − f (0, 0) 0+y 2 0
(0, 0) = lim = lim = lim 3 = lim 0 = 0 ∈ R .
∂y x→0 x−0 y→0 y − 0 y→0 y y→0

This means that f has a partial derivative at (0, 0) with respect to y.

We follow now the steps of the Algorithm from Section 4 of Seminar 6. We
are currently at Step 2. Recall that if the function were is differentiable at (0, 0),
the differential would be the function

df (0, 0) : R2 → R,

defined by
∂f ∂f
df (0, 0)(h1 , h2 ) = h(h1 , h2 , ), ∇f (0, 0)i = h1 (0, 0)+h2 (0, 0) = h1 ·0+h2 ·0, ∀(h1 , h2 ) ∈ R2 .
∂x ∂y

WE DO NOT KNOW IF f IS DIFFERENTIABLE AT (0, 0), SO

In order to study the differentiability of f at (0, 0) we have to study the limit

   
∂f ∂f
f (0, 0) + (h1 , h2 ) − f (0, 0) − h1 ∂x (0, 0) + h2 ∂y (0, 0)
l= lim .
(h1 ,h2 )→(0,0) k(h1 , h2 ) − (0, 0)k
In order for the things to be simpler to analyse, we denote by

ω(h1 , h2 )

the value of a function at a point (h1 , h2 ) ∈ R2 generated by the limit that we have to
compute. So
h1 ·h2 (h21 −h22 )
h21 +h22
−0−0 h1 · h2 (h21 − h22 )
ω(h1 , h2 ) = p =  3 .
h21 + h22 p
2 2 2
h1 + h2
We will make use once again of the sandwich theorem. First of all let us recall that by
the classical mean theorem
h2 + h22 1 1
q
h21 · h22 ≤ 1 ⇐⇒ 2 2
≤ |h1 h2 |.
2 h1 + h2 2
Moreover, it is clear that
|h | |h |
p 1 ≤1 and p 2 ≤ 1.
h21 + h22 h21 + h22
This is why
|h1 | |h | h2 − h22 h2 − h22
|ω(h1 , h2 )| = p 2 · p 2 · p1 ≤ p1
h1 + h22 h21 + h22 h21 + h22 h21 + h22

7
 r
h21 1
≤p 2 ≤ h21 · |h1 h2 |.
h1 + h22 2
Due to the fact that r
1
lim h21 · |h1 h2 | = 0,
(h1 ,h2 )→0 2
from the sandwich theorem we have that

lim ω(h1 , h2 ) = 0.
(h1 ,h2 )→0

So, the limit exists and is 0 in Step 2 of the algorithm. This means that the function f
is differentiable at (0, 0)
In conclusion the function f is differentiable on R2 .

2 Partial Derivatives of Composed Functions.

Even tough the general theory is proved for vector spaces of random dimension, we will
just analyze particular cases so, let us begin with the following example
Eample 2.1:
f : R2 → R3






g : R3 → R 2 ∀(x, y, z) ∈ R3


g(x, y, z) = x2 − y + 2yz 2 , z 3 exy ,




F : R3 → R, F (x, y, z) = (f ◦ g)(x, y, z), ∀(x, y, z) ∈ R3 .


We will determine the partial derivatives of F with respect to

x, y and z, respectively, in terms of the partial derivatives of f .
Note that the actual expression of f is unknown, but it is assumed
that it has partial derivatives with respect to all variables at all
the points in its domain.
Let (x, y, z) ∈ R3 be a random point.
Step 1: The partial derivative with respect to x, i.e

∂F ∂(f ◦ g)
(x, y, z) = (x, y, z).
∂x ∂x
Be aware of the fact that g is a vector function, of vector variable, so from its expression,
since its codomain is R2 it may be written as a pair of two real functions of vector variable,
namely
g = (g1 , g2 ), where g1 , g2 : R3 → R

8
and
g1 (x, y, z) = x2 − y + 2yz 2 and g2 (x, y, z) = z 3 exy .
Then
∂(f ◦ g) ∂f
∂g1 ∂f
∂g2
(x, y, z) = g(x, y, z) · (x, y, z)+ g(x, y, z) · (x, y, z).
∂x ∂g1 ∂x ∂g2 ∂x
In exercises, in order not to get too many thing to write, we might skip some details,
and the formula above may be written as

∂F ∂(f ◦ g) ∂f ∂g1 ∂f ∂g2

= (x, y, z) = · + · .
∂x ∂x ∂g1 ∂x ∂g2 ∂x

With a similar reasoning, we deduce the partial derivatives with respect to y and z
respectively,

∂F ∂(f ◦ g) ∂f ∂g1 ∂f ∂g2

= (x, y, z) = · + ·
∂y ∂y ∂g1 ∂y ∂g2 ∂y
and

∂F ∂(f ◦ g) ∂f ∂g1 ∂f ∂g2

= (x, y, z) = · + · .
∂z ∂x ∂g1 ∂z ∂g2 ∂z
We see that we actually need the partial derivatives of g, with respect to all variables.

 
∂g ∂g1 ∂g2
(x, y, z) = 2x, z 3 exy · y = 2x, yz 3 exy



(x, y, z) = (x, y, z),
∂x ∂x ∂x

 
∂g ∂g1 ∂g2
= −1 + 2z 2 , z 3 exy · x = −1 + 2z 2 , xz 3 exy



(x, y, z) = (x, y, z), (x, y, z)
∂y ∂y ∂y
 
∂g ∂g1 ∂g2
= 4yz, 3z 2 exy .


(x, y, z) = (x, y, z), (x, y, z)
∂z ∂z ∂z
Coming back to F we conclude that

∂F ∂f ∂f
(x, y, z) = (g(x, y, z)) · 2x + (g(x, y, z) · yz 3 exy
∂x ∂g1 ∂g2
∂f ∂f
x2 − y + 2yz 2 , z 3 exy + yz 3 exy x2 − y + 2yz 2 , z 3 exy



= 2x
∂g1 ∂g2

9
 ∂F ∂f ∂f
(x, y, z) = (g(x, y, z)) · (−1 + 2z 2 ) + (g(x, y, z) · xz 3 exy
∂y ∂g1 ∂g2
∂f ∂f
= (−1 + 2z 2 ) x2 − y + 2yz 2 , z 3 exy + xz 3 exy x2 − y + 2yz 2 , z 3 exy



∂g1 ∂g2

∂F ∂f ∂f
(x, y, z) = (g(x, y, z)) · 4yz + (g(x, y, z) · 3z 2 exy
∂z ∂g1 ∂g2
∂f ∂f
x2 − y + 2yz 2 , z 3 exy + 3z 2 exy x2 − y + 2yz 2 , z 3 exy



= 4yz
∂g1 ∂g2
∂f
Here ∂g 1
means the partial derivative of f with respect to the
∂f
first variable, while ∂g2
means the partial derivative of f with
respect to the second variable.
Sometimes, f is written as f (u, v). Pay attention, it is not f = (u, v), and instead of
∂f
∂g1
we may write
∂f
∂u
∂f
, and, instead of ∂g2
we may write
∂f
∂v
.

Example 2.2:
f : R3 → R






g : R2 → R3 ∀(x, y) ∈ R2


g(x, y) = −3x + 2y, x2 + y 2 , 2x3 − y 3 ,




F : R2 → R, F (x, y) = (f ◦ g)(x, y), ∀(x, y) ∈ R2 .


We will determine the partial derivatives of F with respect to

x and y, respectively, in terms of the partial derivatives of f .
Note that the actual expression of f is unknown, but it is as-
sumed that it has partial derivatives with respect to all variables
at all the points in its domain.
Let (x, y) ∈ R2 be a random point.

10
 Step 1: The partial derivative with respect to x, i.e
∂F ∂(f ◦ g)
(x, y) = (x, y).
∂x ∂x
Be aware of the fact that g is a vector function, of vector variable, so from its expression,
since its codomain is R2 it may be written as a pair of two real functions of vector variable,
namely
g = (g1 , g2 , g3 ), where g1 , g2 , g3 : R2 → R
and

g1 (x, y) = −3x + 2y, g2 (x, y) = x2 + y 2 and g3 (x, y) = 2x3 − y 3 .

Then
∂(f ◦ g) ∂f
∂g1 ∂f
∂g2 ∂f
∂g3
(x, y) = g(x, y, z) · (x, y)+ g(x, y) · (x, y)+ g(x, y) · (x, y).
∂x ∂g1 ∂x ∂g2 ∂x ∂g3 ∂x
In exercises, in order not to get too many thing to write, we might skip some details,
and the formula above may be written as

∂F ∂(f ◦ g) ∂f ∂g1 ∂f ∂g2 ∂f ∂g3

= = · + · + · .
∂x ∂x ∂g1 ∂x ∂g2 ∂x ∂g3 ∂x

With a similar reasoning, we deduce the partial derivatives with respect to y and z
respectively,

∂F ∂(f ◦ g) ∂f ∂g1 ∂f ∂g2 ∂f ∂g3

= = · + · + ·
∂y ∂y ∂g1 ∂y ∂g2 ∂y ∂g3 ∂y

We see that we actually need the partial derivatives of g, with respect to all variables.
 
∂g ∂g1 ∂g2 ∂g3
(x, y) = −3, 2x, 6x2


(x, y) = (x, y), (x, y),
∂x ∂x ∂x ∂x
 
∂g ∂g1 ∂g2 ∂g3
(x, y) = 2, 2y, −3y 2


(x, y) = (x, y), (x, y),
∂y ∂y ∂y ∂y
Coming back to F we conclude that
∂F ∂f ∂f ∂f
(x, y) = (g(x, y)) · (−3) + (g(x, y) · 2x + (g(x, y) · 6x2
∂x ∂g1 ∂g2 ∂g3
∂f ∂f
−3x + 2y, x2 + y 2 , 2x3 − y 3 + 2x −3x + 2y, x2 + y 2 , 2x3 − y 3



= −3
∂g1 ∂g2
∂f
+6x2 −3x + 2y, x2 + y 2 , 2x3 − y 3 .


∂g3

11
 ∂F ∂f ∂f ∂f
(x, y) = (g(x, y)) · (2) + (g(x, y) · 2y + (g(x, y) · (−3y 2 )
∂y ∂g1 ∂g2 ∂g3
∂f ∂f
−3x + 2y, x2 + y 2 , 2x3 − y 3 + 2y −3x + 2y, x2 + y 2 , 2x3 − y 3



=2
∂g1 ∂g2
∂f
−3y 2 −3x + 2y, x2 + y 2 , 2x3 − y 3 .


∂g3
∂f
Here ∂g 1
means the partial derivative of f with respect to the
∂f
first variable, while ∂g2
means the partial derivative of f with re-
∂f
spect to the second variable, and ∂g 3
means the partial derivative
of f with respect to the second variable.
Sometimes, f is written as f (u, v, w). Pay attention, it is not f = (u, v, w), and instead
∂f
of we may write
∂g1
∂f
∂u
∂f
instead of ∂g2
we may write
∂f
∂v
∂f
and instead of ∂g3
we may write
∂f
∂w
.

Example 2.3: Let f : R3 → R2 be a differentiable function on

R3 , and let F : R2 → R2 defined by

F (x, y) = f (cos x + sin x, sin x + cos x, ex−y )

 
1 3 4
Knowing that J(f )(1, 1, 1) = determine dF ( π2 , π2 ) .
2 −1 3
Solution:
This exercise once again deals with the differentials of composed functions. First of
all we have to know good all the entry data. Thus:

f : R3 → R2


F : R2 → R2


2 3

 g:R →R g(x, y) = (cos x + sin x, sin x + cos x, ex−y )
F = (f ◦ g)


12
In order to solve the exercise we must acknowledge a result from the lectu-
re, which states that the Jacobi matrix of composed differentiable functions
satisfies the following equality

J(f ◦ g)(a) = J(f )(g(a)) · J(g)(a), (3)

where a is in the domain of g.
We have to do a detective work now :), but fortunately for
us, we have all the date
we need. Thus, the conclusion asks as to determine J(F ) π2 , π2 , so our point is
π π 
a= , .
2 2
We notice that g π2 , π2 = (1, 1, 1). In order to fill in the expressions in (3),

π π  π π  π π 
J(f ◦ g) , = J(f )(1, 1, 1) · J(f ◦ g) , = J(f )(1, 1, ) · J(g)( , .
2 2 2 2 2 2
Since J(f )(1, 1, 1) is already given, we just need J(g) π2 , π2 .

Lucky for us, we have already determined this Martrix in Seminar 6, the last exercise.
(for details, take a look at Seminar6 , just note that the function g was there denoted by
f .)
The searched for Jacobi matrix is
 
π π  −1 0
J(g) , =  0 −1  .
2 2
1 −1

Finally we are able to conclude that

 
π π  π π   1 3 4  −1 0  
3 −7
J(f ◦ g) , · J(f ◦ g) , = ·  0 −1  = .
2 2 2 2 2 −1 3 1 −2
1 −1

Having this Jacobi matrix, we are
able to write the expression of the differential function
associated to F at the point π2 , π2 , namely
π π 
d(F ) , : R2 → R2
2 2

π π   π π   h   3 −7   h   3h − 7h 
1 1 1 2
d(F ) , (h1 , h2 ) = J(F ) , · = · = ,
2 2 2 2 h2 1 −2 h2 h1 − 2h2

for all (h1 , h2 ) ∈ R2 , differential written in its matrix form. In the vector form we have
that π π 
d(F ) , (h1 , h2 ) = (3h1 − 7h2 , h1 − 2h2 ) ∀(h1 , h2 ) ∈ R2 .
2 2

13
 Example 2.4: Let us consider the following:

 f : (0, ∞) × (0, ∞)
→ R a differentiable function
π

g : (0, ∞) × 0, 2 → (0, ∞) × (0, ∞),


 g(ρ, θ) = (ρ cos θ, ρ
sin θ)
F : (0, ∞) × 0, π2 → R F = (f ◦ g)


Determine the partial derivatives of F in terms of the partial

derivatives of f . - Homework

14