IR is continuous in

Chapter 6

DIFFERENTIABLE MAPPINGS

In this chapter we will generalize the notion of derivability known from high school for several variable functions. The notion of differentiability which we will study here will allow us to approximate the value of a function in a certain point with the value of a polynomial of first degree in that point, improving the known results from the continuous functions.

We remind that if f : A IRp _→ IR is continuous in aA, then there exists a neighbourhood UVa for which f

 

x  f(a) for

A U

x  (approximation with a constant). Here we will show that, in some conditions, there is a first-degree polynomial of p variables T1 for which f

 

x T1

 

x , xUA.

1. PARTIAL DERIVATIVES

Definition 1. Let AIRp. A point aA is named interior point

of the set A if there exists a sphere with the centre in a included in A, i.e.

0 r

 such that S

 

a,r  A.

The set of the interior points of the set A is named the interior of the set Aand is denoted Å or Int A.

Remark 1. The set A IRp is open if and only if A=Å. Indeed, if z is open, then for any aA there exists r0 so that S

 

a,r  A, hence a is an interior point and consequently AÅ; from definition 1 results that ÅA, hence A=Å. Conversely, if Å=A, then for any aA there exists r0 so that S

 

a,r  A, hence A is an open set.

Remark 2. The interior of the set A is the largest open set included in A; therefore if G is an open set included in A then GÅ.

Then Å={(x,y,z)  IR3 x,y,z > 0, x2+y2+z2 < 1}.

Proposition 1. If A IRp then Å A'.

Proof. Let aÅ. According to definition 1 results that there

exists r>0 so that S

 

a,r  A; then S

 

a,r A\

 

a Ø, hencea is an accumulation point of the set A.

Corollary. If aÅ then there exists

 

x_n  A\

 

a so that a

xn n 

lim . This affirmation results from proposition 1 and proposition 5 from chapter 5.

Definition 2. Let f : A  IRpIRq be a function and



p



p e,...,e

B  1 respectively Bq 



e1 ,...,eq



the canonical bases of the

linear spaces IRp, respectively IRq. We say that the function f is partially derivable according to xi (where i



1,2,...,p



) in the point

aÅif exists:

 





f(a)] )

te [f(a

t i

IR t t

1 lim

0 IR

q_{. (*)}

This limit is named the partial derivative of the function f

according to (with respect to, or shortly, w.r.t) xiin the point a and is denoted

 

a x

f

i

 

or f_x_i

 

a .

If f has a partial derivative according to every variable we say that f

is partially derivable in a. If A=Å and f is partially derivable according to xiin any point aA we say that f is partially derivable according to xi (on A); in this case the function:

(a) f , a :A

f x

f

i

i x

x i

 

  





q

IR

is named partial derivative of the function f according to xi. If f is continuous on A, partially derivable according to xi in any point aA we say that f is of class C1 on A and we write f _C1(A)

.

Remark 2. In case

 



f



a t



f

 

a



t

a x f , p

t  

   



1 lim 1

0 , or,

denoting x=a+t,

 

 

 

a x

a f x f a

x f

a

x 

 

 



lim ;

then in the case of the functions with a single variable the partial derivative coincides with the derivative and we will keep the notations from high-school:

 

 

a

dx df a

f  . In case p=2, if

 

a,b Å IR2:

 





   

 

 



 

 _t f( a,b t , ) f a,b

a,b f

t

x 10

1 lim

0





 





t a x t _t f at,b  f a,b 

1 lim

0

 

 

a x

a,b f x,b f

a

x 





lim

and, analogously

 

 

 

b y

a,b f a,y f a,b

f

b y y

  





lim .

Remark 3. If f 



f1,...,f_q



, and aÅ, then according to proposition 1, a 



a₁,...,a_p



A' , and from theorem 7 (chapter 5) it results that f is partially derivable if and only if the q scalar components have partial derivatives in a; also:

 

 

 

_

  

 

  

  



a x f ,..., a x f a x

f

i q

i i

1 _.

If we denote xi atei then t0 if and only if xi ai and:

 



 



i i

p j

p i i i j

a x i

j

a x

,...,a a f ,...,a ,a ,x ,...,a a f a

x f

i

i 

 



 _{ }



1 1

1 1

lim .

Example 2. The function: derivable, because:

we say f is partially derivable two times according to xi and xj, hence exists:





 

_

  

 

  



 _x a te f a

f

t i j

t

1 lim

0 IR

q_. We denote this limit (if exists) with

 

a

fx_ix_j or

 

a

x x

f

j i

 2

and we call it the partial derivative of second order (derivative II) of

the function f according xi and xj in a. If ij then

 

a x x

f

j i

 2

and

 

a x x

f

i j

 2

are called mixed derivatives according to xi and xj in a. If i=j we call the derivative II according to xi and xj the partial

derivative of second order (or derivative II) according to xi in a and we denote

 

a x

f

i

2 2

 

, or f

 

a

i

x2 .

Analogously we define the derivatives of higher order. If A=Å, f is continuous and admits partial derivatives of any order hn, continuous on A we say f is of class Cn on A and we write

 

A C f _ n

.

If f _Cn

 

A

for any nIN we say f is of classConA and we write f C

 

A .

The partial derivative of order n according to xi (if exists) is denoted

n i n

x f

 

, or  n

xin

f ,

and the partial derivative of order m according to xj (if exists) for the function  n

xin

f we denote:

m j n i

m n

x x

f

 

 

or n m

x x jm

n i

Remark. For the single variable function derivable n times it understand function u (respectively v). As the partial derivatives are, actually, the derivatives of the partial functions, therefore derivatives of the single variable functions, results that the Leibniz formula is valid for products of scalar functions of several variables as well.

Example 4. Let us compute _{7 6}

Example 5. Let us calculate the mixed derivatives of second

and



,



2 0

0 fy x y    , hence

0

C

f_y ( IR2). But for mixed derivatives

 





_

_

f

 

x y

m m m mx

x f y

x

f _yx

mx y x xy

x xy

mx y

x lim ,

1 3 6 1 ,

lim ,

lim

0 3

2 4 2

0

0   

   

 

  



 ,

hence doesn’t exist limit in origin, namely (0,0) is a point of discontinuity of the mixed derivatives; consequently 2( 2)

IR

C

f  .

As following, we will show that the functions of class C2 have the mixed derivatives equal. For simplifying the notations we will show this property for p2 and q1 in general hypothesis.

Theorem 1. (Schwarz) Let f : A  IR2 IR and

 

a,b Å. If there exists an open neighbourhood VV(a,b), there exists the partial derivatives fx

 

x,y ,fy

 

x,y , fxy

 

x,y for any

 

x,y V, and fxy is

continuous in

 

a,b , then there exists fyx

 

a,b

'' _and

 

a b f

 

a b fxy ,  yx , .

Proof. Let h, k IR* so that



a  h,b  k



V and

 

h k f



a h b k

 

f a h b

 

f a b k

  

f a b

F ,   ,    ,  ,   , (1)

If we denote g

 

x,y  f



x,yk

  

 f x,y , then:

 

h k g



a h b

  

g a b F ,   ,  , .

With respect to the definition of function g it results that:

 

h k hkf



a h b k



F ,  xy  ,  (2)

But f_xy is continuous in (a,b), hence the function

 

h,k  fxy



ah,bk



 fxy

 

a,b

 (3)

 

, 0 lim

0

,k h k 

h  . (4)

Let

 

h

 

h k

k ,

lim 0





 . (5)

Certainly, from (4) and (5) results that:

 

0 lim

0 

 h

h  . (6)

Multiplying the equality (1) with k

1

, replacing F

 

h,k from (2) and



a h b k



fxy  ,  form (3) we obtain:

 



 







 













 f a h b k f a h b

k k h h b a f

h xy , ,

1 ,

, 





 



f a b k f a b



k , ,

1 _{ }

 (7)

Because on V there exists the partial derivative fy , passing to the

limit when k 0 in (7) and considering (5) we have:

 

a b h

 

h f



a h b



f

 

a b f

h xy ,    y  ,  y , ,

or:





 



f a h b f a h



f

   

a b h h y  ,  y ,  xy , 

1

. (8) Passing at the limit in (8) for h0, from (6) it results that fyx

 

a,b exists, and fyx

 

a,b  fxy

 

a,b .

Corollary. Let A  IRp be an open set. If f _C2

 

A

2. DIRECTIONAL DERIVATIVES

We have noticed that the partial derivative of a function w.r.t. the variable xi assumes the existence of the limit (*) (definition 2). In this context the vector

i

te

a tends to a IRp when t0. For example, in IR2, for the derivative in

 

a,b w.r.t the variable x,



at,b



tends (when t0) to

 

a,b along a parallel line with Ox. Let there be (u,v)

 

0,0 . Then the point of coordinates



atu,btv



tends to the point of coordinates

 

a,b (when t0) along the line parallel with the vector ui vj. We

will generalize the notion of partial derivative replacing the vector e_i from (*) with an arbitrary vector, introducing the concept of

(directional) derivative along a vector, concept that plays an important role in the study of electromagnetic phenomena.

Definition 4. Let there be f : A  IRp IRq a vectorial function,



A

a and v IRp\{0} a vector. If there exists



  





q

t

t _t f a tv f a R

R

I I

 





1 lim

0 (**)

we say that f is derivable at a along the vector v. The limit (**), if exists, is denoted

 

a

v f

 

and is called the directional derivative of function f along the vector (with respect to (w.r.t.)) v at the point a. The directional derivative at a of the function f along the versor of the vector v (i.e. with respect to v

v

1

) is also called the (directional) derivative of f in the directionv.

0

(a,b)

0 x

(u,v)

(a,b) y

A

Remark 1. If ve_iB_p then

 

 

a x

f a e

f

i

i 

  



.

Remark 2. For p=q=1, denoting atvx or



x a



v

t 1  , we remark that (**) becomes:

 

   

vf

 

a a

x a f x f a

v f

a

x   

 

 



lim ,

so, in this particular case, the derivability along a vector implies the derivability.

Conversely, if f is derivable in a, and v IR*, taking vtxa we have:

 

 

 







 



 

a

v f v v a f tv a f t a

x a f x f a

f

I t t a

x 

  

 

  



 

1 1 1

lim lim

0

R

, so, derivability implies derivability along any non-zero vector.

Example 6. The derivative along any non-zero vector v of a

linear application is exactly the respective application calculated in v. Truly if fL(IRp, IRq), a IRp and v IRp \{0} then



a tv



f

 

a tf

 

v

f    , for any t IR. From (**) results

 

a f

 

v v

f _

 

, for any a IRp.

Definition 5. Let there be x = (x1,…, xp), y = (y1,…, yp)  IRp. The real number





   

 p

k

p p k

ky x y x y

x y

x

1

1 1 ... ,

is named the scalar product of the vectors x and y.

Remark. The scalar product has the following properties:

(P1) x,x x 2,x IRp and  x,x0x0

(P2)

<

x,y

>=<

y,x>

,



x,y

 IRp

(P3) x,y  x,y,IR,x,y IRp

(P4)  x y,z x,z y,z,x,y,z IRp

Theorem 2 (Riesz). Let fL(IRp, IR). Then there exists a unique a IRp such that:

(a) f

 

x  x,a, for any x IRp

(b) A  a , where A is the matrix of the linear application f.

Proof. Let x



x1,...,x_p



 IR

p

. Then:

 

_

_

 

 

  

   

 

 p

i

i i p

i i

ie x f e x a

x f x f

1 1

, , where a



f

 

e1 ,..., f

 

e_p



 IR

p

. If exists b



b1,...,b_p



 IR

p

such that: f

 

x x,b, for any x IRp, then e_i,be_i,a, for any

p

i1, , so bi ai,i 1,p hence ba and the point (a) is proved. Of course A=a, so (b) is, also, proved.

We will give a set of sufficient conditions for derivability w.r.t. a vector and the formula of the derivative w.r.t. a vector.

Theorem 3(Sufficient conditions for derivability).

Let f : A  IRp IRq and aÅ. If exists a neighbourhood V

V

_{a such that f has partial derivatives on V, continuous in a, then f}

is derivable w.r.t. any vector v IRp\{0} and:

 

_

 

 

  

 p

i

i i

v a x

f a

v f

1

, where v = (v1,…, vp).

Proof. Let t IR* such that atvV. Then:



atv



 f

 

a  f



a tv a tv a_p tv_p

 

 f a a tv a_p tv_p





f _{1 1}, _{2 2},..., ₁, _{2 2},...,



  

 

  



 

 f a1,a2 tv2,a3 tv3,...,ap tvp f a1,a2,a3 tv3,...,ap tvp ...



a a ap ap tvp

 

f a a ap ap



f 1, 2,..., 1,   1, 2,..., 1,

 (1)

Knowing that f has partial derivatives on V, applying the theorem of Lagrange on the intervals



ak,ak tvk



, k 1,p, for the p differences from (1) it results the existence of the elements

  

t a a tv



k p





 



 

 





From the hypothesis of the theorem we know that the partial derivatives are continuous in a; multiplying both members from (2) with

so f is derivable w.r.t. the vector v, and the formula of the directional derivative is proved.

Definition 6. Let f : A IRp IR be a scalar function, partial

is called the gradient of function f, or, the nabla operator applied to f.

Example 7. Let f : IR2_→ IR2,

 



x z x z



ye xe y x

f ,   ,  . Let us calculate the derivative of f in (0,0) w.r.t. a versor v=(v1,v2) which forms an angle of 30 with the axis Ox.

Then

2 3 30 cos 1  

v and

2 1 30 sin 2  

v .

Therefore _

  

  

2 1 , 2

3

v . Moreover:

  





x y x y



x x y x e ye

f ,  1  ,  , f_y

 

x,y 



xexy,



1 y



exy



and, according to theorem 3:

 

 

 

v

 

 

v

y f v x f v

f _{  }

   

  



1 , 0 2 1 0 , 1 2

3 0

, 0 0

, 0 0

,

0 1 2 .

Remark. The derivability of a function in a point w.r.t. any non-zero vector implies the continuity of it along any line which passes through the respective point. Nevertheless it does not imply the continuity for p1.

Example 8. Let f : IR2_→ IR,

 

    

0 ,

2

y x y x f

0 ,

0 ,

 

y y

.

The function f is not continue in (0,0) because





m mx x f

x

1 ,

lim 2

0 

 for



m IR*, so it does not even have limit (w.r.t. the set of variables) in the origin. However it is derivable w.r.t. any vector (u,v)IR2\{(0,0)}, because forv0:

   





 









v u tv tu f t f

v u t f

t t

t

2

0

0 ,

1 lim 0

, 0 ,

0 , 0 1

lim    



 ,

and for v0:

   



,0 0,0



0

1 lim

0  

 _t f tu f

Proposition 2. Let f, g : A IRp IRq and h:A IR

derivable in aÅ w.r.t. v IRp \{0}. Then the functions f g and hf are derivable in a w.r.t. v and



_{ }

_{ }

_{ }

a v g a

v f a

v g f

      

 

 

 

,

          

a h a v

f a f a v h a v hf

   

  



.

Proof. Passing to limit when t0 in the equalities:





 

 











 





g



a tv

  

g a



t a f tv a f t a g f tv a g f

t          

1 1

1

 

  



and

 

   









  









f



a tv



f

 

a



h

 

a t

tv a f a h tv a h t a hf tv a hf

t           

1 1

1

as f



a tv



f

 

a

t0  

lim , according to the previous Remark, we obtain the conclusions from the text.

Remark. The set of functions f : A  IRp IRq derivable in aÅ w.r.t. v with the function addition operation and with the multiplication with scalars from the field IR is, according to proposition 2, a linear space over IR.

Definition 7. Let f : A  IRp IRq,



A

a and two vectors u,v IRp\{0}. If there exist a sphere S

 

a,r  A, the directional derivative (x)

v f

 

for all xS(a,r), and the function v f

 

is derivable w.r.t. the vector u at the point a, we denote this derivative:

 

a u v

f

 

2

, or

 

a v

f

2 2

 

if uv

and we name it derivative of order two of the function f in (at) the point a w.r.t. the vectors v, u, and we say that f is two times derivable in the point a w.r.t. v and u.

Of course, if v e_i and ue_j, then

 

 

a x x

f a

e e

f

j i j

i  

  



2 2

In the same way are defined the higher order derivatives along to the non-zero vectors from IRp.

Remark 1. If v  u 1 then vi cosi,ui cosi,i1,p are the director cosines of the versors v and u i.e.:

p pe

e

vcos1 1...cos and ucos1e1 ...cospep,

and, from theorem 3:

 

_

 

 

and, according to Schwarz theorem:



According to the previous formula:

3. DIFFERENTIABILITY

We will now extend the notion of differentiability studied on functions with a single variable. We remind that the function



A

f : IR_→ IR

is differentiable in aÅ if it exists a linear application daf the differential of the function f in a so that:

 

x  f

 

a d f



xa

 

 xa



(x),

 

a 0

f a  

where :A IR is a continuous function in a. Moreover, f is differentiable in a if and only if f is derivable in a, and

 

h f

 

a h f

da  ' , or using the notations from definition 6,

 

h f

 

a ,h.

f

da  

Definition 8. Let f : A IRp IRq and aÅ. A linear application d_a f : IRp_→ IRq is called the differential of the function f

in (at) the point a (in Fréchet's sense) if:

 

 







  



 



 0

1

lim f x f a d f x a a

x a

a

x IR

q

, or, equivalent (denoting xah)





 

 





0

1 lim

0    

 _h f a h f a daf h

h .

In this case we say that f is differentiable in (at) a. If f is differentiable in any, aA the function f is called differentiable function (on A).

Remark. As in the case of pq1, we can rewrite this definition: the application daf L( IR

p

, IRq) is the differential of the function f in a if it exists a function :A IRqcontinuous in a with

 

a 0

 and

 

x f

 

a d f



x a



x a

 

x

f   a    

or equivalent, it exists a neighbourhood V

V

_{0 and a function} 

V

: 0

 IRq, continuous in 0 IRq so that:



a h



f

 

a d f

 

h h 0

 

h ,(h V)

The use of the expression “the differential of the function f in a” imposes proving the next Proposition.

Proposition 3. If the function f :AIRp_→ IRq is

differentiable in aÅ then its differential in this point is unique.

Proof. Let us suppose that da f and d'a f L( IR

p

, IRq) are differentials of the function f in a. Then:

 

 









 _h d a f h daf h

h '

1 lim

0





 

 







 

 





'



0

1 lim

0        



 _h f a h f a daf h f a h f a d a f h h

(1) Choosing in (1) htek,t0, where



e1,e2,...,ep



is the canonical basis

in IRp we have:

 

 



a k a k



a

 

k a

 

k

o

t _t d f te d f te d f e d f e



 ' '

1 lim

0 ,

meaning

 

 

, 1, . ' f e d f e k p d a k  a k 

So the two linear applications are identical: d'a f da f.

Example 10. Let f : IRp  IRq be a linear application and a IRp. Then



    





0

1 lim

0    

 _h f a h f a f h

h ,

so f is differentiable and daf  f , for any a IR

p

. It follows that any linear application is differentiable and its differential is the linear application itself.

In particular, if g : A  IRp  IRq is differentiable in aÅ and because dagL( IR

p

, IRq) it follows that: db

 

dag dag,b IR

p

. We now try to find a formula for d_a f . We saw that if pq1,

questions we will start by presenting some necessary conditions of differentiability of a function in a point and also the formula for the differential in a point.

Theorem 4. If f : A  IRp  IRq is differentiable in aÅ then: (a) f is continuous in a,

(b) f is derivable along any vIRp\{0} and

 

a d f

 

v.

v f

a

  

(c)

 



 

 



 p

k

k k

a a h

x f h

f d

1

, for any h



h1,....,h_p



 IR

p.

Proof. Let da f L( IR

p

, IRq) be the differential of the function f in the point a.

(a) Because d_a f is the differential of f in a it follows that it exists :A IRq, continuous in a, with 

 

a 0 and

f

 

x  f

 

a daf



xa



 xa

 

x,xA.

But daf is a uniform continuous function (see exercise 24 chap.5), and d_af

 

0 0 so, computing the limit when xa in this equality we obtain f

 

x f

 

a

a

x 

lim ; it follows that f is continuous in a.

(b) Let v  IRp\{0}. From theorem’s hypothesis it follows that:





 

 





0

1 lim

0    

 _h f a h f a daf h

h . (1)

Choosing htv, with t IR*, because h0 IRp it follows that

0



t , and from (1) we obtain:





 

 





0,

1 lim

0     

 _{t v} f a tv f a da f tv

t

or, equivalent:



  

 









  



 

   



 _{  }

 

  



 _{t t} f a tv f a d f v

t v

f td a f tv a f

t a t a

t

1 lim 1

lim 0

0

0 ,

from where:



  





 

0

1 lim

0    

 _t f a tv f a da f v

t . (2)



  

function

 

h k  f

 

h f

 

k k 

 

h k 

L , 'x 0,0 'y 0,0 , , IR

2

.

must be the differential of the function f in (0,0). Let us study the differentiability of f in a with the definition 8:

   



   

   



but not sufficient, as shown in the examples that follow:

 

, 0

 

0,0. lim

0

,y f x y f

x   

Also f is partial derivable in (0,0) and:

 

0,0 lim

   

,0 0,0 0

 

0,0.

0 y

x

x f

x f x f

f     



If f would be differentiable in the origin, using the theorem 4, then the linear function:

 

,

 

0,0

 

0,0 0

   



 y

y f x x f y x L

must be the differential of the function f in (0,0). But f is not differentiable in (0,0) because the function

         





 

_{2 2}

2

2 ,

1 ,

0 , 0 ,

, 1 ,

y x

xy y

x f y x y

x L f

y x f y x y x

  

 

 

 ,

   

x,y  0,0, does not have a limit in (0,0).

Prove using another method that f is not differentiable in the origin!

Example 13. The function

 

   

  

0 y , 0

0 y y x y x, f

2

,

is derivable along any vector (not null), but it is not differentiable in (0,0).

Truly, if w =(u,v) IR2\{(0,0)}

 

0,0 lim1





,



 

0,0



, 2

0 _v

u f

tv tu f t w

f

t  

 





for v0 and for v0:

 

 



,0 0,0



0. 1

lim ) 0 , 0 (

0  

 



 _t f tu f

w f

t

On the other hand,



2



2 0 ,

lim f mx x m

x 

so f is not continuous in (0,0); by following the point (a) from theorem 4, the function f is not differentiable in (0,0).

Remark. If A= Å and f is differentiable on A then f C0

 

A . We now give sufficient conditions of differentiability.

Theorem 5. Let f : A  IRp IRqand aÅ. If it exists V

V

_{a so}

that f admits partial derivatives on V continuous in a, then f is differentiable in a.

Proof. We will show, using theorem 4 (c), that demonstration of theorem 3 replacing tv with h. More precise, from the existence of partial derivatives on the interval



a,ah



V , applying Lagrange's theorem, it follows the existence of the elements:

 



,..., , , ,...,





,...,



0

1

1 1

1 1

1 1

1 _ 

   

 

 





    

p

k

p k

p p k k k k k k k

a a x

f h a h a h c h a h a x

f

when h0, because

k

x f

 

is continuous in a, k 1,p. It follows that

 

0 lim

0 

 h

h  , f is differentiable in a and

 





_

 

p

k

k x

af h f a h

d _k

1

' , for any







 h h_p

h 1,..., IR p

.

Corollary. If f _C1

 

A

then f is differentiable on A.

Example 14. Let us compute d_{ }0,0 f

 

1,1 for the function

f : IR2 IR2, f

 

x,y 



x2  y,exy



. Because

 





 



x y



y y x

x x y x e f x y e

f ,  2 ,  ,  ,  1,  ,

   

0,0  0,1



x

f , fy

   

0,0  1,1 , and

1

C

f  ( IR2), it follows that f is differentiable on IR2 so also in (0,0) and:

 0,0

 

,

 

0,0

      

0,0v 0,u v,v v,u v



,

y f u x f v u f

d    

   

 

and d_{ }0,0 f

   

1,1  1,2 .

Remark. The sufficient conditions of differentiability given at theorem 5 are not also necessary, as shown by the following example.

Example 15. The function

 

   

 



0

1 sin ) (

, 2 2

2 2

y x y

x y x

f

0 ,

0 ,

2 2

2 2

 

 

y x

y x

is differentiable in

 

0,0. Truly, because:

 

0,0 lim

   

,0 0,0 lim sin1 0

 

0,0, 0

0 x y

x

x f

x x x

f x f

f      

 

 

, 1

 

, 2 2 sin ₂ 1 ₂ 0, 2

2 _    



y x y x y x f y x y x



when x,y0, so f is differentiable in (0,0). On the other hand 0

C

f  ( IR2), admits partial derivatives on IR2 but they are not continuous in (0,0). Indeed

 

, 2 sin ₂ 1 _{2 2}2 ₂ cos ₂ 1 ₂ ,

y x y x

x y

x x y x fx

 

  

 for

   

x,y  0,0

and for the sequence _

  

 

   

 



 2 n

1 , n 2

1

which tends to (0,0) (when

 

n ),  

  

 

 



 n n

n

fx 2

2 1 , 2

1

, so, according to Heine's theorem, f_x is not continuous in (0,0).

Remark 1. We are now able to synthesize some connexions

between the notions that we discussed. For an interior point of the domain A of the function f the following implications take place:

and none of these implications are, in general, equivalences.

Remark 2. In computing differentials we use for the

projections prk : IR p

→ IR the notation dxk,k 1,p. We remind that







 

 _{k p}

p

k h h h h h h

dx ( 1,..., ) , 1,..., IR p

.

In the case pq1 pr₁ dx and:

 

a dx L dx

df f

da   ( IR, IR).

In the case p 2 and q1, pr1 dx and pr2 dy and, according to

theorem 4(c): Differentiability

Continuity Relative continuity with respect to the

sets {a+tv|tIR}

Partial continuity Derivability along any

not null vector v

 

 

 

a b dy L

y f dx b a x f f

dab 

   



 , ,

, ( IR

2

, IR),

formula, which represents the differential as a linear application. If

 

x,y IR2then

d(a,b)f(x,y) = (a,b) y y

f (a,b) x x

f _

   



IR,

formula which represents the differential of the function f in (a,b)

computed in (x,y). If f is differentiable in A = Å  IR2 we will use the total differential of f:

df = dy

y f dx x f

   



and the differential operator:

d = dy

y dx

x 

  



,

operator that associates to a differentiable function its total differential df. The previous notations and formulas can be extended for any p, q  IN*:

- the differential of f in the point a:

da f = (a) dx L x

f

k p

k k

 





1

( IRp, IRq)

- the differential of f in the point a calculated in x=(x1, …,xp): daf (x) =



 



 k

p

k

x a x

f _k

1

' _IRq

- the total differential of f on an open set:

d f = k

p

k k

dx x

f



 



1

- differential operator:

d = k

p

k k

dx x



 



1

.

Let us now consider that f =



f1,..., fq



: A  IR

p _

daf (x) = i

which is called the Jacobi matrix (or Jacobian matrix) of function f in the point a. Then:

The differential formula shows us that some of the properties of the operators operator.

=g

 

a daf  f

 

a dag,

for any aA, hence d

 

fg = fdggdf .

(b) da (f+g)=

 

x

 

k p

k

x a g a dx

f _{k k} )

( 1

  







 =

 p

 

_k

k

x a dx

f _k







1

+

 

_k

p

k

x a dx

g _k







1

=daf dag, for any aA, therefore d



f g



=df dg.

(c) _

    

g f

da =

 



   

   

x



k p

k

x a g a f a g a dx

f a

g k  k



1 2

1

=

=

 

a g2

1



g(a)

 

_k

p

k

x a dx

f _k







1

-f(a)

 

_k

p

k

x a dx

g _k







1



₌

=

 

a



g

 

a d f f

 

a d g



g2 a  a

1

,

for any aA, therefore _

    

g f

d =



gdf fdg



g2 

1

.

Geometrical Interpretations

1. We observed that the function f : A  IR_→ IR is differentiable in aÅ if and only if f is derivable in a, therefore the curve of equation y= f

 

x , xA (which is the graph of the function f) admits a tangent in the point M



a, f

 

a



of equation

y-f(a) = f'

 

a xa



, or, equivalent

y= f

 

a da f



xa



.

2. If f : A= Å  IR2  IR, fC1

 

A _and

 

A b

a,  then the equation:

C : f(x,y) – f(a,b) = 0

is represented geometrically as a curve which goes through the point P(a,b). Let us suppose that this equation can be explicitly given with respect to x, this means there is a function g : B IR IR

such that

Then:

 

x,y -g

 

x

fx   , and fy

 

x,y 1;

the equation

 f



xa yb



0

da,b ,

is equivalent with

 

, 



 

 

, 



-g

 





  0

 a b x a f a b y b a x a y b

fx y ,

or

 

a x a



b

y g 

Therefore the equation

0 ) , ( ) ,

( f xa yb 

d ab

is the equation tangent to the curve C in the point P with the slope m=g

 

a = -

 

_{ }

b a f

b a f

y x

, ,

 

, and grad f

 

a,b = -g

 

a i  j is a perpendicular vector tothis line.

3. If f : A IR3  IR, f_C1

 

A _and





A c b

a, ,  then the equation:

S: f



x,y,z



 f



a,b,c



is represented graphically by a plane which goes through P(a,b,c) and the equation:



,



0

) ,

( f xa yb 

d ab

x O

y=g(x)

P ) b , a ( f



 , , f



xa,yb,zc



0

dabc

or equivalent:

: f_x



a,b,c



xa



 f_y



a,b,c



yb



 f_z



a,b,c



zc



=0 is represented graphically by a plane which goes through P, is tangent to S and its norm is the vector:

grad f(a,b,c) =fx



a,b,c



i fy



a,b,c



j fz



a,b,c



k .

4. Let us now consider f : A IR2  IR, fC1

 

A . Then the graphic of the function f is a surface of equation

S : z = f(x,y), (x,y) A,

surface that admits a tangent plane in P0



a,b, f

 

a,b





 

a,b A



of equation:

: z-f(a,b) =

 







 

a x b a x f

,

 

a b y b



y

f _

 

, =d_{ }a,b f



xa,yb



Let there P



x,y, f

 

x,y



S and PP’ the line which goes through P, parallel to Oz. Then Q(x, y, f(a,b) +d_{ }a,b f(xa,yb))

PP’, and the distance from P to Q is:

d(P,Q)= |(f

 

x,y - f

 

a,b -d_{ }a,b f



xa,yb



|. Therefore the approximate relation:

f

 

x,y  f

 

a,b +d_{ }a,b f



xa,yb



0

x

z

S s

)

P

(

f



y

is equivalent to d(P,Q)0; therefore, if P is ''close enough to'' P0,

then the point PS is ''close'' to the point Q   and the value of f(x,y) is approximately equal with the value in (x-a, y-b) of the 1st grade polynomial f(a,b) + d(a,b)f.

Example 16. Let f : IR2  IR, f(x,y) =

4 2

2 2

y x _

. The equation f(x,y) = f(1, 2 ), or the equivalent:

4 2

2 2

y x _

= 1 is represented graphically by an ellipse with the semi axes 2 and 2, and the equation:

) 2 , 1 (

d f(x-1, y- 2) = 0,

is equivalent to 2 (x –1) + 2 (y - 2 )= 0, or to the equation x  y = 2  1

and is represented graphically by a line with the slope m = -1 tangent to the ellipse in P(1, 2 ); the vector

O

x

Q P0

P S

z



P’(x,y,0)

grad f(1, 2 ) = 2 ((i  j) is perpendicular to this line.

Example 17. The equation f(x, y, z) = f ( , 3

3 2 , 3 1

), where f (x, y, z) = x2 +

9 4

2 2

z y _

is represented graphically by an ellipsoid of the equation x2 +

9 4

2 2

z y _

=1 (of semi axes 1, 2 and 3), and the equation:

0 3 , 3 2 , 3 1 3

, 3 2 , 3

1 

  



 _{  }

   



 x y z

d f , or



3



0

3 3

2 3 2 3

1 3 1 3

2 _{  }

  

   

   



 x y z ,

x



1

,

2



f



represents a plane tangent to the ellipsoid in P _

  

 

3 , 3 2 , 3 1

;

grad f( k

3 3

2 j 3 1 i 3 2 3 , 3 2 , 3

1 _{  }

   

 

is a vector perpendicular to this plane.

Example 18. If

  





 

,

    

, f ñ,ó x,y f : 0,  0,2 IR IR- 00  ,





x

where cos and y=sin, are the functions which make the transition to the polar coordinates, then:





_

  

  

  

   



cos sin

sin cos

,

f

J , and

 



,





,

D y x D

.

Of course f is invertible and f 1

  

x,y  ,



_{, where}

2

y



 x

 and

x y

óArctan . Then:

 

    

 

    

 

 



 





2 2 2

2

2 2 2

2

1

y x

x y x

y

y x

y y x

x

x,y

J_f ,

and

 

_{ }





ñ

y x

y x x,y

D

ñ,ó

D 1

3 2 2

2 2

 



 .

4. THE DIFFERENTIALS AND DERIVATIVES OF

COMPOSITE FUNCTIONS

Theorem 6. If the function f : A IRp _→B IRq is differentiable in aÅ, and g : B IRr is differentiable in b=f(a)



B, then g_{f : A} _IRr _{is differentiable in a and:}



g f



da  = df(a)gdaf .

Proof. Because f is differentiable in aÅ there exists  : A→

IRq such that:

 

0 lim 

a x

x  and f

 

x  f

 

a daf



xa



 xa á

 

x (1)

Because g is differentiable in b=f (a)



B there exists  : B→ IRr such that:

 

0

lim 

b x

x  and g

 

y g

 

b dbg



yb



 yb 

 

y (2)

By putting y=f(x) in (2), from (1) it follows that: g_{f (x)= g}_{f(a)+ d g}

b



daf



xa



 xa 

 

x



+

+ daf



xa



 xa 

 

x 

 

y = gf (a)+ +dbgdaf



xa



 xa 

 

x , where for x a :

 

x

 =

a x

1



_

_{ }

_

_

_

_{   }



y x a x a x f d x g d a

x b   a      . (3)

We must also prove that a x lim

 

 

x =0. Using the triangle inequality in

(3) we obtain:

 

x

  dbg((

 

x ) + a x

1

) y (





daf



xa



 xa  (x)



. (4)

But daf  L (IRp,IRq) and dbg  L ( IRq,IRr); using the inequality form exercise 14, chapter 5, it follows that:



x a



J

 

a x a f