of 45 /45
1. Folosind deﬁnit ¸ia s˘ a se calculeze derivatele part ¸iale f 0 x (x 0 ,y 0 ), f 0 y (x 0 ,y 0 ) ˆ ın punctele indicate (x 0 ,y 0 ), pentru urm˘ atoarele funct ¸ii f : D R 2 R, unde D este domeniul maxim de derivabilitate : 1. f (x, y)= x 3 y + x, (x 0 ,y 0 ) = (1, 1); 2. f (x, y) = ln(x 2 + y 3 ), (x 0 ,y 0 )=(-1, 1); 3. f (x, y)= e 3x+y + e x+3y , (x 0 ,y 0 ) = (1, -2); 4. f (x, y)= 3 p x 5 y 2 , (x 0 ,y 0 ) = (1, -1); 5. f (x, y) = arctan xy, (x 0 ,y 0 ) = (2, 1); 6. f (x, y)= r xy + x y , (x 0 ,y 0 ) = (2, 1). Rezolvare: 1. f 0 x (1, 1) = lim x1 f (x, 1) - f (1, 1) x - 1 = lim x1 x 3 + x - 2 x - 1 =4, f 0 y (1, 1) = lim y1 f (1,y) - f (1, 1) y - 1 = lim y1 y - 1 y - 1 =1 2. f 0 x (-1, 1) = lim x→-1 f (x, 1) - f (-1, 1) x +1 = lim x→-1 ln(x 2 + 1) - ln 2 x +1 = = lim x→-1 2x x 2 +1 = -1 f 0 y (-1, 1) = lim y1 f (-1,y) - f (-1, 1) y - 1 = lim y1 ln(1 + y 3 ) - ln 2 y - 1 = = lim y1 3y 2 1+ y 3 = 3 2 2. S˘ a se studieze existent ¸a derivatelor part ¸iale, continuitatea ¸ si diferent ¸ia- bilitatea ˆ ın origine pentru funct ¸iile f : R 2 R, deﬁnite prin: 1. f (x, y)= 3xy x 2 + y 2 , dac˘ a (x, y) 6= (0, 0) 0, dac˘ a (x, y) = (0, 0); 2. f (x, y)= xy p x 2 + y 2 , dac˘ a (x, y) 6= (0, 0) 0, dac˘ a (x, y) = (0, 0); 3. f (x, y)= xy x 2 - y 2 x 2 + y 2 , dac˘ a (x, y) 6= (0, 0) 0, dac˘ a (x, y) = (0, 0); 1

# derivate partiale

Embed Size (px)

### Text of derivate partiale

1. Folosind definitia sa se calculeze derivatele partiale f ′x(x0, y0), f′y(x0, y0)

ın punctele indicate (x0, y0), pentru urmatoarele functii f : D ⊂ R2 → R,unde D este domeniul maxim de derivabilitate :

1. f(x, y) = x3y + x, (x0, y0) = (1, 1);

2. f(x, y) = ln(x2 + y3), (x0, y0) = (−1, 1);

3. f(x, y) = e3x+y + ex+3y, (x0, y0) = (1,−2);

4. f(x, y) = 3√x5y2, (x0, y0) = (1,−1);

5. f(x, y) = arctan xy, (x0, y0) = (2, 1);

6. f(x, y) =

√xy +

x

y, (x0, y0) = (2, 1).

Rezolvare:

1. f ′x(1, 1) = limx→1

f(x, 1)− f(1, 1)

x− 1= lim

x→1

x3 + x− 2

x− 1= 4,

f ′y(1, 1) = limy→1

f(1, y)− f(1, 1)

y − 1= lim

y→1

y − 1

y − 1= 1

2. f ′x(−1, 1) = limx→−1

f(x, 1)− f(−1, 1)

x+ 1= lim

x→−1

ln(x2 + 1)− ln 2

x+ 1=

= limx→−1

2x

x2 + 1= −1

f ′y(−1, 1) = limy→1

f(−1, y)− f(−1, 1)

y − 1= lim

y→1

ln(1 + y3)− ln 2

y − 1=

= limy→1

3y2

1 + y3=

3

2

2. Sa se studieze existenta derivatelor partiale, continuitatea si diferentia-bilitatea ın origine pentru functiile f : R2 → R, definite prin:

1. f(x, y) =

3xy

x2 + y2, daca (x, y) 6= (0, 0)

0, daca (x, y) = (0, 0);

2. f(x, y) =

xy√x2 + y2

, daca (x, y) 6= (0, 0)

0, daca (x, y) = (0, 0);

3. f(x, y) =

xyx2 − y2

x2 + y2, daca (x, y) 6= (0, 0)

0, daca (x, y) = (0, 0);

1

4. f(x, y) =

(x2 + y2) cos1

x2 + y2, daca (x, y) 6= (0, 0)

0, daca (x, y) = (0, 0).

Rezolvare:

1. f ′x(0, 0) = limx→0

f(x, 0)− f(0, 0)

x= 0,

f ′y(0, 0) = limy→0

f(0, y)− f(0, 0)

y= 0.

Deoarece limx→0,y=mx

f(x, y) = limx→0

3mx2

(1 +m2)x2=

3m

1 +m2, m ∈ R, rezulta

ca f nu este continua ın origine.Daca f ar fi diferentiabila ın origine, atunci ar rezulta ca f este continua ın

origine, ceea ce este absurd. Prin urmare, f nu este continua si diferentiabilaın origine, dar admite derivate partiale ın origine.

2. f ′x(0, 0) = 0, f ′y(0, 0) = 0.

Deoarece |f(x, y| = |xy|√x2 + y2

≤ |xy||x|

= |y|, avem lim(x,y)→(0,0)

f(x, y) = 0 =

f(0, 0), deci functia f este continua ın origine.Daca f ar fi diferentiabila ın origine, atunci ar exista

df(0, 0) = f ′x(0, 0) + f ′y(0, 0) = 0 · dx+ 0 · dy = 0

si lim(x,y)→(0,0)

f(x, y)− f(0, 0)− df(0, 0)(x, y)√x2 + y2

= lim(x,y)→(0,0)

xy

x2 + y2= 0. Dar,

cum limx→0,y=mx

xy

x2 + y2=

m

1 +m2, m ∈ R, limita de mai sus nu exista. Asadar

f nu este diferentiabila ın origine.

3. Sa se determine derivatele partiale de ordinul ıntai si diferentiala deordinul ıntai pentru urmatoarele functii f : D ⊂ R2 → R, unde D estedomeniul maxim de diferentiabilitate:

1. f(x, y) =√x2 + y2;

2. f(x, y) = 2x2y − 3xy + x2 − y + 2;

3. f(x, y) = lnxy;

4. f(x, y) = ln yx;

5. f(x, y) = ln yy − lnxx;

6. f(x, y) = lnx

y;

2

7. f(x, y) = x ln y − ey2+x;

8. f(x, y) = xexy;

9. f(x, y) = ex2−y;

10. f(x, y) = xy;

11. f(x, y) = ex2+y2 sin2 x;

12. f(x, y) = arctanx

y;

13. f(x, y) = arctan√x2 + y2;

14. f(x, y) = x2 arctan(x2 + y2);

15. f(x, y) = arcsiny

x, x > 0;

16. f(x, y) =√

1− x2 − y2;

17. f(x, y) = ln(x+√x2 + y2);

18. f(x, y) =x2 − y2

x2 + y2;

19. f(x, y) = x2 − 3xy + 2y2 + 3x− 4y + 2;

20. f(x, y) = sin(2x+ 3y);

21. f(x, y) = x3 + y3 + 3xy;

22. f(x, y) = (x− a)(y − a)(x− b)(y − b), unde a, b ∈ R;

23. f(x, y) = xy2ex−y;

24. f(x, y) = x4y3(1− x2 + y);

25. f(x, y) = cos x+ cos y + sin(x+ y);

26. f(x, y) = sinx sin y sin(x+ y);

27. f(x, y) = ln 3√x2 + y2 + arctanxy;

28. f(x, y) = lnm n√x2 + y, unde m ∈ N∗, n ∈ N, n ≥ 2;

29. f(x, y) = (x2 − y2) siny

x;

3

30. f(x, y) = sin2(x2y + xy3);

31. f(x, y) =x

y+y

x;

32. f(x, y) =xy3

x4 + y2;

33. f(x, y) = sinx

x2 + y2;

34. f(x, y) = (x2 + y2) cos1√

x2 + y2;

35. f(x, y) =1

x+x

y+

1

y.

4. Sa se determine derivatele partiale de ordinul ıntai si diferentiala deordinul ıntai pentru urmatoarele functii f : D ⊂ R3 → R, unde D estedomeniul maxim de diferentiabilitate:

1. f(x, y, z) = (x+ y) sin yz;

2. f(x, y, z) = x2 + y2 + xyz;

3. f(x, y, z) = 3x2 + 2y2 + 3z2 + 2xy − 3yz;

4. f(x, y, z) = −x3y2z + 10xy − 2z3 + 4;

5. f(x, y, z) =√x2 + y2 + z2;

6. f(x, y, z) =x√

y2 + z2;

7. f(x, y, z) = arctanxy

z;

8. f(x, y, z) = arctanz√

x2 + y2;

9. f(x, y, z) = lnz

xy;

10. f(x, y, z) = ln xyyzzx;

11. f(x, y, z) = lny√

x2 + z2;

4

12. f(x, y, z) = zex−y2

+ yex+z;

13. f(x, y, z) = xy + yz + zx;

14. f(x, y, z) = exyz3

cos(x+ y);

15. f(x, y, z) =x

y− y

z+z

x;

16. f(x, y, z) = (x+ a)(y + b)(z + c), unde a, b, c ∈ R;

17. f(x, y, z) = zexy;

18. f(x, y, z) = zex2+y2 + xyez;

19. f(x, y, z) = sinx cos y sin(x+ z);

20. f(x, y, z) = eyx (z2 − x2);

21. f(x, y, z) = xy2z3(6− 2x+ 3y − 4z);

22. f(x, y, z) = xzx2 − y2

x2 + z2;

23. f(x, y, z) =(x− y)n

(y − z)m, unde n,m ∈ N∗;

24. f(x, y, z) =y2

x+z

y− 2

z;

25. f(x, y, z) = cos2(ax+ by + cz), unde a, b, c ∈ R;

5. Sa se determine derivatele partiale de ordinul al doilea si diferentialade ordinul al doilea ın punctele indicate (x0, y0) pentru functiile f : D ⊂R2 → R, unde D este domeniul maxim de diferentiabilitate:

1. f(x, y) =√x2 + y2, (x0, y0) = (1,−1);

2. f(x, y) = exy, (x0, y0) = (1, 0);

3. f(x, y) = ln x2y, (x0, y0) = (3, 1);

4. f(x, y) = lnx

y2, (x0, y0) = (1, 2);

5. f(x, y) = arctan(x2 + y2), (x0, y0) = (−1, 1);

6. f(x, y) = (1 + x)m(1 + y)n, m, n ≥ 2, (x0, y0) = (0, 0);

5

7. f(x, y) = ex cos y, (x0, y0) = (0, π);

8. f(x, y) =x+ y

x− y, (x0, y0) = (2, 1);

9. f(x, y) = x3 + y3 − x2 − y2, (x0, y0) = (1, 1);

10. f(x, y) = ex2−y2 , (x0, y0) = (1,−1).

Rezolvare:

1. f ′x(x, y) =x√

x2 + y2, f ′y(x, y) =

y√x2 + y2

,

f ′′x2(x, y) =y2

(x2 + y2)√x2 + y2

, f ′′xy(x, y) = − xy

(x2 + y2)√x2 + y2

,

f ′′y2(x, y) =x2

(x2 + y2)√x2 + y2

,

f ′′x2(1,−1) = f ′′y2(1,−1) = f ′′xy(1,−1) =1

2√

2,

d2f(1,−1) =1

2√

2(dx2 + 2dxdy + dy2) =

1

2√

2(dx+ dy)2

2. f ′x(x, y) = yexy, f ′y(x, y) = xexy, f ′′x2(x, y) = y2exy, f ′′y2(x, y) =

x2exy, f ′′xy(x, y) = (1 + xy)exy, f ′′x2(1, 0) = 0, f ′′y2(1, 0) = 1, f ′′xy(1, 0) =

1, d2f(1, 0) = 2dxdy + dy2

6. Sa se determine derivatele partiale de ordinul al doilea si diferentialade ordinul al doilea ın punctele indicate (x0, y0, z0) pentru functiile f : D ⊂R3 → R, unde D este domeniul maxim de diferentiabilitate:

1. f(x, y, z) = ln(1 + x+ y + z), (x0, y0, z0) = (1, 0, 1);

2. f(x, y, z) =1√

x2 + y2 + z2, (x0, y0, z0) = (1, 1, 1);

3. f(x, y, z) = xeyz + yezx + zexy, (x0, y0, z0) = (1, 1, 1);

4. f(x, y, z) = x6yz4 + x4y3− yz2 + 6x+ 2yz − 2, (x0, y0, z0) = (1, 2,−1);

5. f(x, y, z) = zexyz, (x0, y0, z0) = (1, 1, 1);

6. f(x, y, z) = x arctany

z, (x0, y0, z0) = (2, 1, 1);

7. f(x, y, z) = sin2(x− y − z), (x0, y0, z0) = (π,π

2,π

3);

6

8. f(x, y, z) =x+ z

xy + z, (x0, y0, z0) = (1,−1, 2);

9. f(x, y, z) =√

9− x2 − y2 − z2, (x0, y0, z0) = (2, 1, 1);

10. f(x, y, z) = 3

√x− zy − z

, (x0, y0, z0) = (2, 0, 1) .

Rezolvare:

1. f ′x(x, y, z) = f ′y(x, y, z) = f ′z(x, y, z) =1

1 + x+ y + z

f ′′x2(x, y, z) = f ′′y2(x, y, z) = f ′′z2(x, y, z) = − 1

(1 + x+ y + z)2

f ′′xy(x, y, z) = f ′′xz(x, y, z) = f ′′yz(x, y, z) = − 1

(1 + x+ y + z)2

d2f(1, 0, 1) = −1

9(dx2 +dy2 +dz2 + 2dxdy+ 2dxdz+ 2dydz) = −1

9(dx+dy+

dz)2.

7. Sa se determine derivatele partiale de ordinul al doilea si diferentiala deordinul al doilea ıntr-un punct curent (x, y) pentru functiile f : D ⊂ R2 → R,unde D este domeniul maxim de diferentiabilitate:

1. f(x, y) = x3y2 − 2xy3 + 3x2 − 6y + 7;

2. f(x, y) = x4 + y4 − 2x2 + 5xy − 3y6;

3. f(x, y) =2x− yx+ 3y

;

4. f(x, y) = ex−y2;

5. f(x, y) = yexy;

6. f(x, y) = (x2 + y2)ex+y;

7. f(x, y) = ln(x2 + y3);

8. f(x, y) = sinxy;

9. f(x, y) = arctanx

y;

10. f(x, y) = yx.

7

Rezolvare:

1. f ′x(x, y) = 3x2y2 − 2y3 + 6x, f ′y(x, y) = 2x3y − 6xy2 − 6,f ′′x2(x, y) = 6xy2 + 6, f ′′y2(x, y) = 2x3 − 12xy, f ′′xy(x, y) = 6x2y − 6y2,

d2f(x, y) = 6(xy2 + 1)dx2 + 12(x2y − y2)dxdy + 2(x3 − 6xy)dy2.

8. Sa se determine derivatele partiale de ordinul al doilea si diferentialade ordinul al doilea ıntr-un punct curent (x, y, z) pentru functiile f : D ⊂R3 → R, unde D este domeniul maxim de diferentiabilitate:

1. f(x, y, z) =x2 + y2 + z2

xyz;

2. f(x, y, z) =x+ z

y + z;

3. f(x, y, z) = xy2z3(6− x+ 2y − z);

4. f(x, y, z) = z2 − xey + yez + zex;

5. f(x, y, z) = (x+ z)eyx ;

6. f(x, y, z) = y√x2 + z;

7. f(x, y, z) = ln(x+ y + z);

8. f(x, y, z) = ln xyz;

9. f(x, y, z) = cos(x+ 2y + 3z);

10. f(x, y, z) = x sin y + y sin z + z sinx.

Rezolvare:

1. f ′x(x, y, z) =1

yz− y

x2z− z

x2y, f ′y(x, y, z) = − x

y2z+

1

xz− z

xy2,

f ′z(x, y, z) = − x

yz2− y

xz2+

1

xy, f ′′x2(x, y, z) =

2(y2 + z2)

x3yz,

f ′′y2(x, y, z) =2(x2 + z2)

xy3z, f ′′z2(x, y, z) =

2(x2 + y2)

xyz3,

f ′′xy(x, y, z) =−x2 − y2 + z2

x2y2z, f ′′xz(x, y, z) =

−x2 + y2 − z2

x2yz2

f ′′yz(x, y, z) =x2 − y2 − z2

xy2z2,

8

d2f(x, y, z) =2(y2 + z2)

x3yzdx2 +

2(x2 + z2)

xy3zdy2 +

2(x2 + y2)

xyz3dz2 +

+2(−x2 − y2 + z2)

x2y2zdxdy +

2(−x2 + y2 − z2)x2yz2

dxdz +2(x2 − y2 − z2)

xy2z2dydz

9. Sa se scrie matricea Jacobi si sa se determine jacobianul pentruurmatoarele functii:

1. f : (0,∞)× [0, 2π]→ R2, f(r, t) = (r cos t, r sin t);

2. f : (0,∞)× [0, 2π]× [−π2,π

2]→ R3,

f(r, θ, ϕ) = (r cos θ cosϕ, r sin θ cosϕ, r sinϕ);

3. f : R2 → R2, f(x, y) = (xex−y, x3y+x2−xy+ y+ 3), ın punctul (1,1).

Rezolvare:

1. Fie f1(r, t) = r cos t si f2(r, t) = r sin t.

Atunci Jf (r, t) =

((f ′1)r(r, t) (f ′1)t(r, t)(f ′2)r(r, t) (f ′2)t(r, t)

)=

(cos t −r sin tsin t r cos t

),

si detJf (r, t) =

∣∣∣∣cos t −r sin tsin t r cos t

∣∣∣∣ = r.

10. Sa se scrie matricea Jacobi pentru urmatoarele functii:

1. f : R2 → R3, f(x, y) = (xy, x+ y, x− y);

2. f : (R∗)3 → R2, f(x, y, z) =

(xyz,

x

yz

);

3. f : (R∗)3 → R3, f(x, y, z) =

(a

x+ yz, bx+

1

y− 1

z

), a, b ∈ R.

11. Sa se determine gradientul pentru urmatoarele functii f : D → R,unde D ⊂ Rn, n ∈ {2, 3} este domeniul maxim de diferentiabilitate:

1. f(x, y) = lnx

y;

2. f(x, y) = (x2 + y2)ex−y;

3. f(x, y) = y ln(x2 + y2);

4. f(x, y) = x3y2 + 3x2y − 2y + 7, ın punctul (1,-1);

9

5. f(x, y, z) = arctanz

xy;

6. f(x, y, z) = sin(ax+ by + cz), a, b, c ∈ R;

7. f(x, y, z) = x2 + y2 + z2, ın punctul (-1,1,2).

Rezolvare:

1. gradf (x, y) = (f ′x(x, y), f ′y(x, y)) =

(1

x,−1

y

).

2. gradf (x, y) = (x2 + y2 + 2x,−x2 − y2 + 2y)ex−y.

12. Sa se calculeze derivata urmatoarelor functii dupa directia si puncteleindicate:

1. f : R2 → R, f(x, y) = x2y − xy2, dupa directia

(1

2,

√3

2

), ın punctul

M(1,-1);

2. f : R2 → R, f(x, y) = ye−x+y, dupa directia axei Ox, ın punctulM(1,1);

3. f : R2 → R, f(x, y) = 2x3 + 3xy − 3y2 + 2, dupa directia gradf (1, 1),ın punctul M(-1,1);

4. f : R2 → R, f(x, y) =√x2 + y2, dupa directia gradientului sau ıntr-un

punct curent M(x,y), x2 + y2 6= 0;

5. f : R∗ × R → R, f(x, y) = arctany

x, dupa directia vectorului MN , ın

punctul M(1,1), unde N(3,2);

6. f : R3 → R, f(x, y, z) = sinhx+cosh y+sinh z, dupa directia vectoruluiMN , ın punctul M(1,2,-1), unde N(-1,1,1);

7. f : R3 → R, f(x, y, z) = x2 + y2 + z2, dupa directia gradf (1,−1, 1), ınpunctul M(1,-1,1);

8. f : R3 → R, f(x, y, z) = x3 + y3 + z3 − 3xyz, dupa directia axei Oz, ınpunctul M(1,-1,2);

9. f : R3 → R, f(x, y, z) = ex2+y2 sin2 z, dupa directia

(1√3,

1√3,− 1√

3

),

ın punctul M(

1, 0,π

4

);

10

10. f : R3 → R, f(x, y, z) = xy + yz + zx, dupa directia MN , ın punctulM(1,0,4), unde N(3,-1,3);

Rezolvare:

1. f ′x(x, y) = 2xy − y2, f ′y(x, y) = x2 − 2xy

df

du(1,−1) =

1

2f ′x(1,−1) +

√3

2f ′y(1,−1) =

3(√

3− 1)

2.

10. Deoarece−−→MN = (2,−1,−1), ||

−−→MN || =

√6, iar versorul directiei

−−→MN este u =

(2√6,− 1√

6,− 1√

6

)avem

df

du(1, 0, 4) =

2√6f ′x(1, 0, 4)− 1√

6f ′y(1, 0, 4)− 1√

6f ′z(1, 0, 4) =

√6

3.

0.1 Diferentiabilitatea functiilor compuse

13. Sa se calculeze df(x), daca f = f(u, v), u = u(x), v = v(x), pentruurmatoarele functii:

1. f(u, v) = u2v + 3v − 2, unde u = x3 + x2 + x+ 1, v = x4 − 1;

2. f(u, v) = unvm, unde u = sinx, v = cosx, n,m ∈ N∗;

3. f(u, v) = ln(u+ v2), unde u = ax2 + x+ 2, v = x3− bx2 + a, a, b ∈ R;

4. f(u, v) = uv, unde u = ln2 x, v = tan3(x2 + 1);

5. f(u, v) =u

v, unde u = ex, v = lnx;

6. f(u, v) = uv, unde u = sinx, v = cosx;

7. f(u, v) = ln cosu√v

, unde u = 3x2, v =√x2 + 1.

Rezolvare:

1. f ′(x) = f ′u(u, v)u′(x) + f ′v(u, v)v′(x)

Intrucat f ′u = 2uv, f ′v = u2 + 3, u′(x) = 3x2 + 2x+ 1, v′(x) = 4x3, dupaformula de mai sus rezulta f ′(x) = 2uv(3x2 + 2x+ 1) + 4x3(u2 + 3),df(x) = (2uv(3x2 + 2x+ 1) + 4x3(u2 + 3))dx.

11

3. Deoarece f ′u =1

u+ v2, f ′v =

2v

u+ v2, u′(x) = 2ax+1, v′(x) = 3x2−2bx,

obtinem

f ′(x) =2ax+ 1 + 2v(3x2 − 2bx)

u+ v2, df(x) =

2ax+ 1 + 2v(3x2 − 2bx)

u+ v2dx.

14. Sa se calculeze f ′x(x, y) si f ′y(x, y), daca f = f(u, v), u = u(x, y), v =v(x, y), pentru urmatoarele functii:

1. f(u, v) =√u2 + v2, unde u = x− y, v =

x

y;

2. f(u, v) = arctanv

u, unde u = x2 + y, v = x+ y2;

3. f(u, v) = lnuv, unde u = x+ y, v = xy;

4. f(u, v) = un+vm−(uv)nm, unde u = sin(x−y), v = cosxy, n,m ∈ N∗;

5. f(u, v) = veu2v, unde u = ln(x+ y), v = xy;

6. f(u, v) = arctanu

v, unde u = x sin y, v = x cos y.

Rezolvare:

1. f ′x = f ′uu′x + f ′vv

′x, f

′y = f ′uu

′y + f ′vv

′y.

Cum f ′u =u√

u2 + v2, f ′v =

v√u2 + v2

, u′x(x, y) = 1, u′y(x, y) = −1,

v′x(x, y) =1

y, v′y(x, y) = − x

y2, dupa formulele de mai sus rezulta

f ′x =uy + v

y√u2 + v2

, f ′y = − uy2 + xv

y2√u2 + v2

.

2. Deoarece f ′u = − v

u2 + v2, f ′v =

u

u2 + v2, u′x(x, y) = 2x, u′y(x, y) =

1, v′x(x, y) = 1, v′y(x, y) = 2y, dupa formulele de mai sus rezulta

f ′x =u− 2xv

u2 + v2, f ′y =

2uy − vu2 + v2

.

15. Sa se calculeze df(x, y), respectiv df(x, y, z) pentru urmatoarelefunctii:

1. f(x, y) = ϕ(xy);

12

2. f(x, y) = ϕ(x+ y, x− y);

3. f(x, y) = ϕ(x3y − 5x2 + 2xy2 + 4x+ 1, x2y5 − 7y);

4. f(x, y) = ϕ(xy,

y

x

);

5. f(x, y, z) = ϕ(xyz, x2 − y2 + z2);

6. f(x, y, z) = ϕ(x+ yz, ex2y+z, lnxyz);

7. f(x, y, z) = ϕ(√

x2 + y2 + z2, x arctany

z

);

8. f(x, y, z) = ϕ(e−x2+yz, xeyz, exyz).

Rezolvare:

1. Fie u = xy. Astfel f(x, y) = ϕ(u(x, y)).Avem

f ′x = ϕ′(u)u′x = yϕ′(u), f ′y = ϕ′(u)u′y = xϕ′(u),

df(x, y) = (ydx+ xdy)ϕ′(u).

2. Fie u = x+ y, v = x− y. Astfel f(x, y) = ϕ(u(x, y), v(x, y)).Avem

f ′x = ϕ′uu′x + ϕ′vv

′x = ϕ′u + ϕ′v,

f ′y = ϕ′uu′y + ϕ′vv

′y = ϕ′u − ϕ′v,

df(x, y) = (ϕ′u + ϕ′v)dx+ (ϕ′u − ϕ′v)dy.

16. Sa se calculeze d2f(x, y), respectiv d2f(x, y, z) pentru urmatoarelefunctii:

1. f(x, y) = ϕ(x+ y, x− y);

2. f(x, y) = ϕ

(xy,

x

y

);

3. f(x, y) = ϕ(lnxy, exy);

4. f(x, y, z) = ϕ(x+ y − z, x− y + z,−x+ y + z);

5. f(x, y, z) = ϕ

(xyz,

x

yz

);

6. f(x, y, z) = ϕ(x+ yz, x2 + y2 − z2) .

13

Rezolvare:

1. Daca notam u = x+ y, v = x− y, atunci f(x, y) = ϕ(u(x, y), v(x, y))si

f ′x = ϕ′u + ϕ′v, f′y = ϕ′u − ϕ′v, f ′′x2 = ϕ′′u2 + 2ϕ′′uv + ϕ′′v2 ,

f ′′xy = ϕ′′u2 − ϕ′′v2 , f ′′y2 = ϕ′′u2 − 2ϕ′′uv + ϕ′′v2 ,

d2f(x, y) = (ϕ′′u2 +2ϕ′′uv+ϕ′′v2)dx2 +2(ϕ′′u2−ϕ′′v2)dxdy+(ϕ′′u2−2ϕ′′uv+ϕ′′v2)dy

2.

2. Daca notam u = xy, v = xy, atunci

f ′x = yϕ′u +1

yϕ′v, f

′y = xϕ′u −

x

y2ϕ′v, f

′′x2 = y2ϕ′′u2 + 2ϕ′′uv +

1

y2ϕ′′v2 ,

f ′′xy = ϕ′u −1

y2ϕ′v + xyϕ′′u2 −

x

y3ϕ′′v2 , f

′′y2 = x2ϕ′′u2 −

2x2

y2ϕ′′uv +

x2

y4ϕ′′v2 +

2x

y3ϕ′v,

d2f(x, y) =

(y2ϕ′′u2 + 2ϕ′′uv +

1

y2ϕ′′v2

)dx2 +

2

(ϕ′u −

1

y2ϕ′v + xyϕ′′u2 −

x

y3ϕ′′v2

)dxdy+

(x2ϕ′′u2 −

2x2

y2ϕ′′uv +

x2

y4ϕ′′v2 +

2x

y3ϕ′v

)dy2.

17. Sa se arate ca:

1. f(x, y) = ϕ(ax− by) verifica ecuatia bf ′x + af ′y = 0, a, b ∈ R∗;

2. f(x, y) = ϕ

(x

y

)verifica ecuatia xf ′x + yf ′y = 0;

3. f(x, y) = ϕ(axn − byn) verifica ecuatia byn−1f ′x + axn−1f ′y = 0, a, b ∈R∗, n ∈ N∗;

4. f(x, y) = xyϕ(y2 − x2) verifica ecuatia yf ′x + xf ′y =

(x

y+y

x

)f ;

5. f(x, y) = yϕ

(x

y

)verifica ecuatia xf ′x + yf ′y = f ;

6. f(x, y) = ϕ(√x2 + y2) verifica relatia yf ′x = xf ′y;

7. f(x, y) = ϕ(xy) +√xyψ

(yx

)verifica ecuatia x2f ′′x2 − y2f ′′y2 = 0;

8. f(x, y) = eyϕ

(ye

x2

2y2

)verifica ecuatia (x2 − y2)f ′x + xyf ′y = xyf ;

14

9. f(x, y) = ϕ(x2 + y2) verifica ecuatia yf ′x − xf ′y = 0;

10. f(x, y) = ϕ(lnxy, xy) verifica ecuatia xf ′x − yf ′y = 0;

11. f(x, y, z) = ϕ(xz, x2 − y2 + z2) verifica ecuatiaxyf ′x + (x2 − z2)f ′y − yzf ′z = 0;

12. f(x, y) = ϕ(x+ ay) verifica relatia f ′x = af ′y;

13. f(x, y) = yϕ(x2 − y2) verifica relatia1

xf ′x +

1

yf ′y =

f

y2;

14. f(x, y) = xy + xϕ(yx

)verifica relatia xf ′x + yf ′y = xy + f ;

15. f(x, y) = xϕ(yx

)+ ψ

(yx

)verifica ecuatia x2f ′′x2 + 2xyf ′′xy + y2f ′′y2 = 0.

Rezolvare:

1. Daca u = ax − by, atunci f ′x = aϕ′(u), f ′y = −bϕ′(u), bf ′x + af ′y =baϕ′(u)− abϕ′(u) = 0.

18. Sa se calculeze derivatele partiale de ordinul n pentru urmatoarelefunctii:

1. f(x, y) = eax+by, a, b ∈ R;

2. f(x, y) = sin(ax+ by), a, b ∈ R;

3. f(x, y) = (x+ y)eax+by, a, b ∈ R;

4. f(x, y) = x2ex+y;

5. f(x, y) = ln(x+ y);

6. f(x, y, z) = eax+by+cz, a, b, c ∈ R;

7. f(x, y, z) = cos(ax+ by + cz), a, b, c ∈ R;

8. f(x, y, z) = ln(ax+ by + cz), a, b, c ∈ R .

Rezolvare:

1. f(n)xn = aneax+by, f

(n)yn = bneax+by, f

(n)

xkyn−k = akbn−keax+by.

3. Folosind formula lui Leibniz pentru derivarea de ordin superior obtinem

f(n)xn =

n∑k=0

Ckn(x+ y)(k)(eax+by)(n−k) = (ax+ ay + n)an−1eax+by,

15

f(n)yn = (bx+ by + n)bn−1eax+by,

f(n)

xkyn−k = ak−1bn−k(ax + by + k)eax+by + (n − k)akbn−k−1eax+by = (abx +

aby + kb+ a(n− k))ak−1bn−k−1eax+by.

19. Sa se calculeze:

1. f(n)xn (x, y) si f

(n)yn (x, y), pentru f(x, y) = xy3exy;

2. f(3)xyz(x, y, z), f

(3)

x2y(x, y, z), f(3)

x3 (x, y, z), pentru f(x, y, z) = exyz;

3. f(3)

xz2(x, y, z), f(3)

y3 (x, y, z), f(3)xyz(x, y, z), pentru f(x, y, z) = exy+yz+zx .

0.2 Extreme libere

20. Sa se determine punctele de extrem local pentru urmatoarele functiif : R2 → R:

1. f(x, y) = x2 + y2 − 2x+ 1;

2. f(x, y) = x2 + y2 + 4x− 6y + 2;

3. f(x, y) = x2 + y2 − xy + 2x+ 1;

4. f(x, y) = xy +1

x+

1

y+ 2, x, y ∈ R∗;

5. f(x, y) = x4 + y4 − 2x2 + 4xy − 2y2 + 4;

6. f(x, y) = x3 + y3 − 6xy + 3;

7. f(x, y) = (x+ y2)ex+y;

8. f(x, y) = xy2ex−y, x, y ∈ R∗;

9. f(x, y) = sinx sin y sin(x+ y), x, y ∈ (0, 2π);

10. f(x, y) = (x+ 1)(y + 1)(x+ y);

11. f(x, y) = (x+ 2)2 + (x− 6)2 + (x− 5)2 + y2 + (y + 4)2;

12. f(x, y) = x2y2(1− x− y), x, y ∈ R∗ .

16

Rezolvare:

1. Incepem prin determinarea punctelor stationare, care sunt solutii ale

sistemului

{f ′x(x, y) = 0

f ′y(x, y) = 0. Avem f ′x = 2x − 2, f ′y = 2y. Obtinem punctul

critic a = (1, 0). Pentru a verifica daca a este punct de extrem folosimcriteriul lui Sylvester aplicat matricei hessiene ın acest punct. Intrucat f ′′x2 =

2, f ′′xy = 0, f ′′y2 = 2, atunci Hf (a) =

(f ′′x2(a) f ′′xy(a)f ′′xy(a) f ′′y2(a)

)=

(2 00 2

), iar

∆1 = 2 > 0, ∆2 = 4 > 0, rezulta ca a = (1, 0) este punct de minim local.

6. f ′x = 3x2 − 6y, f ′y = 3y2 − 6x.

{f ′x(x, y) = 0

f ′y(x, y) = 0⇒ (0, 0), (2, 2) puncte

critice.Avem f ′′x2 = 6x, f ′′xy = −6, f ′′y2 = 6y.

Hf (2, 2) =

(12 −6−6 12

); ∆1 = 12 > 0, ∆2 = 108 > 0, deci (2, 2) este

punct de minim local.f ′′x2(0, 0)f ′′y2(0, 0) − (f ′′xy(0, 0))2 = −36 < 0, deci (0, 0) nu este punct de

extrem.

21. Sa se determine punctele de extrem local pentru urmatoarele functiif : R3 → R:

1. f(x, y, z) = (x− a)2 + (y − b)2 + (z − c)2, a, b, c ∈ R;

2. f(x, y, z) = x2 + y2 + z2 − 2x+ 6y − 8z;

3. f(x, y, z) = x+y2

4x+z2

y+

2

z, x, y, z > 0;

4. f(x, y, z) = (x−2)2 + (x+ 4)2 + (y+ 2)2 + (y−4)2 + (z−1)2 + (z+ 1)2;

5. f(x, y, z) = x2 + y2 + z2 + xy − x+ z + 1;

6. f(x, y, z) =1

x+x

y+y

z+

z

16, x, y, z > 0;

7. f(x, y, z) = 2x2 + y2 − xy − xz + 2z;

8. f(x, y, z) = x3 + y2 + z2 + 12xy + 2z, x, y, z ∈ R∗;

9. f(x, y, z) = (x− 1)2 + (y + 2)2 + z2 − 3z + 5;

10. f(x, y, z) = 2x2 − y2 − 4z2 − 4xy − 2x− 2y − 4z − 1;

17

11. f(x, y, z) = x2 + 4y2 + 9z2 + 6xy − 2x .

Rezolvare:

1. Determinam punctele stationare rezolvand sistemul

f ′x(x, y, z) = 0

f ′y(x, y, z) = 0

f ′z(x, y, z) = 0.

,

2(x− a) = 0

2(y − b) = 0

2(z − c) = 0

. Obtinem punctul critic (x, y, z) = (a, b, c), iar ma-

tricea hessiana ın acest punct este de formaHf (a, b, c) =

2 0 00 2 00 0 2

. Intrucat

∆1 = 2 > 0,∆2 =

∣∣∣∣2 00 2

∣∣∣∣ = 4 > 0,∆3 =

∣∣∣∣∣∣2 0 00 2 00 0 2

∣∣∣∣∣∣ = 8 > 0 rezulta (a, b, c)

punct de minim local.

0.3 Functii implicite

22. Pentru functia y = f(x), sa se determine:

1. f ′(1) si f ′′(1), daca f(1) = 1 si x2 + 2xy − y2 + x− y − 2 = 0;

2. f ′(1) si f ′′(1), daca f(1) = −1 si x2 + 2xy − y2 + x+ y + 2 = 0;

3. f ′(0) si f ′′(0), daca f(0) = 1 si x3 + y2 − xy2 + y − x− 2 = 0;

4. f ′(0) si f ′′(0), daca f(0) = −1 si x3 + y3 − x2y2 + y2 + x− y + 1 = 0;

5. f ′(1) si f ′′(1), daca f(1) = 1 si x4 + y4 + xy − 3 = 0.

Rezolvare:

1. Deoarece suntem ın conditiile teoremei functiilor implicite, aplicandaceasta teorema, obtinem

f ′(x) = −F′x(x, y)

F ′y(x, y)= −2x+ 2y + 1

2x− 2y − 1, f ′(1) = 5

f ′′(x) = −(2 + 2f ′(x))(2x− 2y − 1)− (2− 2f ′(x))(2x+ 2y + 1)

(2x− 2y − 1)2,

f ′′(1) = −28.

23. Pentru functia z = f(x, y), sa se determine:

18

1. f ′x(1, 1), f ′y(1, 1), daca f(1,1)=1 si x3 + y3 + z3 + 3xyz − 6 = 0;

2. f ′x(1, 0), f ′y(1, 0), daca f(1,0)=-1 si xy + yz + zx+ 1 = 0;

3. f ′x(0,−1), f ′y(0,−1), daca f(0,-1)=1 si xyz + 2yex − y2z + 3 = 0;

4. f ′x(1,−1), f ′y(1,−1), daca f(1,-1)=1 si x2y + y2z + z2x− 1 = 0;

5. f ′x(2, 2), f ′y(2, 2), daca f(2,2)=0 si (x+ y)ez − xy − z = 0;

6. f ′x(0, 0), f ′y(0, 0), daca f(0,0)=0 si z2 − xey − yez − zex = 0 .

Rezolvare:

1. Deoarece suntem ın conditiile teoremei functiilor implicite, aplicandaceasta teorema, obtinem

f ′x(x, y) = −F′x(x, y, z)

F ′z(x, y, z)= −x

2 + yz

z2 + xy, f ′x(1, 1) = −1,

f ′y(x, y) = −F ′y(x, y, z)

F ′z(x, y, z)= −y

2 + xz

z2 + xy, f ′y(1, 1) = −1.

24. Pentru functia y = f(x), sa se determine derivatele partiale de ordinulıntai si al doilea:

1. ln 3√x2 + y2 − arctan

y

x= 0;

2. x3 + y3 − x+ y − 2 = 0;

3. ey − ex + xy = 0;

4. x2 + y2 + xy + 1 = 0;

Rezolvare:

1. f ′(x) = −F′x(x, y)

F ′y(x, y)=

2x+ 3y

3x− 2y,

f ′′(x) =(2 + 3y′)(3x− 2y)− (3− 2y′)(2x+ 3y)

(3x− 2y)2.

Inlocuind y′ cu f ′(x) determinat mai sus obtinem f ′′(x) =26(x2 + y2)

(3x− 2y)2.

2. f ′(x) = −F′x(x, y)

F ′y(x, y)=

1− 3x2

1 + 3y2,

19

f ′′(x) = −6(x+ y)(1 + 9xy(x2 − xy + y2))− 6xy(x− y)

(1 + 3y2)3.

25. Pentru functia z = f(x, y), sa se determine derivatele partiale deordinul ıntai si al doilea:

1. x2 + 2y2 + z2 − 4z + 8 = 0;

2. x3 + y3 + z3 − 3xyz = 0;

3. x2 + y2 + z2 − a2 = 0 .

Rezolvare:

1. f ′x(x, y) = −F′x(x, y, z)

F ′z(x, y, z)= − x

z − 2, f ′y(x, y) = −

F ′y(x, y, z)

F ′z(x, y, z)= − 2y

z − 2,

f ′′x2(x, y) = −z − 2− xz′z(z − 2)2

= −x2 + (z − 2)2

(z − 2)3, f ′′y2(x, y) = −

2(z − 2− yz′y)(z − 2)2

=

−2(2y2 + (z − 2)2)

(z − 3)3, f ′′xy(x, y) =

xz′y(z − 2)2

= − 2xy

(z − 2)3.

26. Pentru functiile y = f(x) si z = g(x) sa se calculeze :

1. f ′(1) si g′(1), daca f(1) = 1, g(1) = 1 si{x+ 2y − z − 3 = 0

x3 + y3 + z3 − 2xyz − 2 = 0 ;

2. df, dg, d2f, d2g, daca

{xyz = a

x+ y + z = b, a, b ∈ R;

3. f ′(1), g′(1), f ′′(1), g′′(1), daca f(1) = 1, g(1) = 1 si{x2 + y2 − z2 = 1

x2 + 2y2 + 3z2 = 6 ;

4. df, dg, daca

{x2 + y2 + z2 − 1 = 0

x2 + y − z = 0 ;

5. df, dg, d2f, d2g, daca

{x2 + y2 + 3z2 = 1

x2 + y2 − z2 = 0 .

Rezolvare:

1. Fie F (x, y, z) = x+2y−z−3 = 0, G(x, y, z) = x3+y3+z3−2xyz−2 =0.

20

f ′(x) = −

D(F,G)

D(x, z)

D(F,G)

D(y, z)

= −

∣∣∣∣ 1 −13x2 − 2yz 3z2 − 2xy

∣∣∣∣∣∣∣∣ 2 −13y2 − 2xz 3z2 − 2xy

∣∣∣∣ =−3x2 − 3z2 + 2xy + 2yz

3y2 + 6z2 − 4xy − 2xz,

f ′(1) = −2

3.

g′(x) = −

D(F,G)

D(y, x)

D(F,G)

D(y, z)

= −

∣∣∣∣ 2 13y2 − 2xz 3x2 − 2yz

∣∣∣∣∣∣∣∣ 2 −13y2 − 2xz 3z2 − 2xy

∣∣∣∣ =−6x2 + 3y2 − 2xz + 4yz

3y2 + 6z2 − 4xy − 2xz,

g′(1) =1

3.

2. f ′(x) =y(x− z)

x(z − y), g′(x) =

z(y − x)

x(z − y), f ′′(x) = ((z − y)2 + (x − z)2 +

(x− y)2)yz

x2(y − z)3, g′′(x) = −f ′′(x).

27. Sa se arate ca functia z = f(x, y), definita prin F (x−az, y− bz) = 0,unde a, b ∈ R, verifica relatia af ′x(x, y) + bf ′y(x, y) = 1.

Rezolvare:

Fie u = x − az si v = y − bz. Atunci derivand pe rand ın raport cu xrespectiv y obtinem

F ′u(1− af ′x)− bF ′vf ′x = 0

aF ′uf′y + F ′v(1− bf ′y) = 0,

de unde rezulta

f ′x(x, y) =F ′u(u, v)

aF ′u(u, v) + bF ′v(u, v)

si

f ′y(x, y) =F ′v(u, v)

aF ′u(u, v) + bF ′v(u, v).

Astfel

af ′x(x, y) + bf ′y(x, y) =aF ′u(u, v)

aF ′u(u, v) + bF ′v(u, v)+

bF ′v(u, v)

aF ′u(u, v) + bF ′v(u, v)= 1.

28. Sa se arate ca functia z = f(x, y), definita prin:

21

1. (y + z) sin z − y(x+ z) = 0, verifica ecuatia z sin zf ′x − y2f ′y = 0;

2. y(x+ z)− (y+ z)ϕ(z) = 0, verifica ecuatia z(x+ z)f ′x− y(y+ z)f ′y = 0;

3. ϕ(xz,y

z

)= 0, verifica relatia xf ′x + yf ′y = z;

4. x2 +y2 +z2 = ϕ(x+y+z), verifica relatia (y−z)f ′x+(z−x)f ′y = x−y;

5. x2 + y2− 2xz− 2yϕ(z) = 0, verifica ecuatia (y2− x2 + 2xz)f ′x + 2y(z−x)f ′y = 0.

Rezolvare:

1. f ′x =y

sin z + (y + z) cos z − y, f ′y =

x+ z − sin z

sin z + (y + z) cos z − y.

Atunci

z sin zf ′x−y2f ′y = z sin zyz sin z

sin z + (y + z) cos z − y− y2(x+ z − sin z)

sin z + (y + z) cos z − y=

y(z sin z − y(x+ z) + y sin z)

sin z + (y + z) cos z − y= 0.

29. Sa se determine extremele functiei y = f(x), definita prin:

1. x3 + y3 − 3xy = 0;

2. y2 + 2yx2 − 4x− 3 = 0;

3. x2 − 2xy + 5y2 − 2x+ 4y + 1 = 0;

4. x3 + y3 − 3x2y − 3 = 0;

5. y3 + x2 − xy − 3x− y + 4 = 0;

6. (x2 + y2)2 = a2(x2 − y2), a ∈ R∗.

Rezolvare:

1. Punctele critice ale functiei y = f(x) sunt solutiile sistemului

F ′x(x, y) = 0

F (x, y) = 0

F ′y(x, y) 6= 0

.

F ′x = 3x2 − 3y, F ′y = 3y2 − 3x. Rezolvand sistemul

3x2 − 3y = 0

x3 + y3 − 3xy = 0

3y2 − 3x 6= 0

rezulta punctul critic x = 3√

2.

22

Deoarece f ′(x) =y − x2

y2 − x, f ′′(x) =

(y′ − 2x)(y2 − x)− (2yy′ − 1)(y − x2)(y2 − x)2

,

iar f ′′( 3√

2) = −2 < 0, rezulta ca punctul x = 3√

2 este punct de maxim local.

2. Punctele critice ale functiei y = f(x) sunt solutiile sistemului

F ′x(x, y) = 0

F (x, y) = 0

F ′y(x, y) 6= 0

.

F ′x = 4xy−4, F ′y = 2y+2x2. Rezolvand sistemul

xy − 1 = 0

y2 + 2yx2 − 4x− 3 = 0

y + x2 6= 0

rezulta punctul critic x =1

2.

Deoarece

f ′(x) =4(1− xy)

x2 + 2y, f ′′(x) =

4((−y − xy′)(x2 + 2y)− (2x+ 2y′)(1− xy))

(x2 + 2y)2,

iarf ′′(

1

2

)= −8 < 0, rezulta ca punctul x =

1

2este punct de maxim local.

30. Sa se determine extremele functiei z = f(x, y), definita prin:

1. x2 + y2 + z2 − 2x+ 4y − 6z − 11 = 0;

2. x3 − y2 − 3x+ 4y + z2 + z − 8 = 0;

3. x2 + y2 + z2 − xz − yz + 2x+ 2y + 2z − 2 = 0.

Rezolvare:

1. Cautam punctele critice solutii ale sistemului

F ′x(x, y, z) = 0

F ′y(x, y, z) = 0

F ′z(x, y, z) 6= 0

F (x, y, z) = 0

solutiile sistemului

2x− 2 = 0

2y + 4 = 0

2z − 6 6= 0

x2 + y2 + z2 − 2x+ 4y − 6z − 11 = 0

,

care sunt (x, y, z) = (1,−2,−2) respectiv (x, y, z) = (1,−2, 8).Avem

f ′x(x, y) =1− xz − 3

, f ′y(x, y) = −y + 2

z − 3, f ′′x2(x, y) = −z − 3 + (1− x)z′x

(z − 3)2,

f ′′xy(x, y) =(x− 1)z′y(z − 3)2

= 0, f ′′y2(x, y) =−z + 3 + (y + 2)z′y

(z − 3)2.

23

Aplicand criteriul lui Sylvester pentru hessiana ın punctele critice de mai

sus, obtinem Hf (1,−2,−2) =

1

50

01

5

, ∆1 =1

5> 0, ∆2 =

1

25> 0, deci

(1,−2,−2) este punct de minim local, respectivHf (1,−2, 8) =

−1

50

0 −1

5

,

∆1 = −1

5< 0, ∆2 =

1

25> 0, asadar(1,−2, 8) este punct de maxim local.

31. Pentru functiile u = f(x, y) si v = g(x, y) sa se determine:

1. df(0, 0), dg(0, 0), d2f(0, 0), d2g(0, 0), daca f(0, 0) = 1, g(0, 0) = 0 si{x+ y + u+ v − 1 = 0

x2 + y2 + u2 + v2 − 1 = 0 ;

2. df(1, 1), dg(1, 1), d2f(1, 1), d2g(1, 1), daca f(1, 1) = 1, g(1, 1) = 0 si{x+ 2y − u+ v = 2

x3 + y3 + u3 − 3xyu− 3v = 0 ;

3. df(0, 1), dg(0, 1), d2f(0, 1), d2g(0, 1), daca f(0, 1) = 1, g(0, 1) = 1 si{u = x+ y

uv = y ;

Rezolvare:

1.

f ′x = −

D(F,G)

D(x, v)

D(F,G)

D(u, v)

= −

∣∣∣∣ 1 12x 2v

∣∣∣∣∣∣∣∣ 1 12u 2v

∣∣∣∣ =x− vv − u

, f ′x(0, 0) = 0,

f ′y = −

D(F,G)

D(y, v)

D(F,G)

D(u, v)

= −

∣∣∣∣ 1 12y 2v

∣∣∣∣∣∣∣∣ 1 12u 2v

∣∣∣∣ =y − vv − u

, f ′y(0, 0) = 0, df(0, 0) = 0,

g′x = −

D(F,G)

D(u, x)

D(F,G)

D(u, v)

= −

∣∣∣∣ 1 12u 2x

∣∣∣∣∣∣∣∣ 1 12u 2v

∣∣∣∣ =u− xv − u

, g′x(0, 0) = −1,

24

g′y = −

D(F,G)

D(u, y)

D(F,G)

D(u, v)

= −

∣∣∣∣ 1 12u 2y

∣∣∣∣∣∣∣∣ 1 12u 2v

∣∣∣∣ =u− yv − u

, g′y(0, 0) = −1, dg(0, 0) = −dx− dy,

f ′′x2 =(−v′x + 1)(v − u)− (v′x − u′x)(x− v)

(v − u)2, f ′′x2(0, 0) = −2,

f ′′xy =−v′y(v − u)− (x− v)(v′y − u′y)

(v − u)2, f ′′xy(0, 0) = −1,

f ′′y2 =(1− v′y)(v − u)− (v′y − u′y)(y − v)

(v − u)2, f ′′y2(0, 0) = −2,

d2f(0, 0) = −2(dx2 + dxdy + dy2),

g′′x2 =(u′x − 1)(v − u)− (v′x − u′x)(u− x)

(v − u)2, g′′x2(0, 0) = 2,

g′′xy =u′y(v − u)− (v′y − u′y)(u− x)

(v − u)2, g′′xy(0, 0) = 1,

g′′y2 =(u′y − 1)(v − u)− (v′y − u′y)(u− y)

(v − u)2, g′′y2(0, 0) = 2,

d2g(0, 0) = 2(dx2 + dxdy + dy2).

32. Pentru functiile u = f(x, y) si v = g(x, y) sa se determine derivatelepartiale de ordinul ıntai, daca:

1. u+ v = x+ y, xu+ yv = 1;

2. u+ v = x, u− yv = 0;

3. x+ y + u+ v = a, x3 + y3 + u3 + v3 = b, a, b ∈ R;

4. x+ y = u+ v, y sinu = x sin v.

Rezolvare:

1.

f ′x = −

D(F,G)

D(x, v)

D(F,G)

D(u, v)

= −

∣∣∣∣−1 1u y

∣∣∣∣∣∣∣∣1 1x y

∣∣∣∣ =y + u

y − x,

25

f ′y = −

D(F,G)

D(y, v)

D(F,G)

D(u, v)

= −

∣∣∣∣−1 1v y

∣∣∣∣∣∣∣∣1 1x y

∣∣∣∣ =y + v

y − x,

g′x = −

D(F,G)

D(u, x)

D(F,G)

D(u, v)

= −

∣∣∣∣1 −1x u

∣∣∣∣∣∣∣∣1 1x y

∣∣∣∣ =−u− xy − x

,

g′y = −

D(F,G)

D(u, y)

D(F,G)

D(u, v)

= −

∣∣∣∣1 −1x v

∣∣∣∣∣∣∣∣1 1x y

∣∣∣∣ =−v − xy − x

.

0.4 Extreme cu legaturi

33. Sa se determine punctele de extrem ale urmatoarelor functii f : R2 → R,cu legaturile specificate:

1. f(x, y) = x2 + y2, 2x+ 3y = 1 ;

2. f(x, y) = xy, −x+ y = 1;

3. f(x, y) = x2 + y2,x

a+y

b− 1 = 0, a, b ∈ R∗;

4. f(x, y) = 2x+ y, x2 + y2 = 5;

5. f(x, y) = 3x− y, x2 − y2 = 2;

6. f(x, y) = xy, x+ y = 1;

7. f(x, y) = x2 + y2 − 2y + 1, y2 − x2 = 1;

8. f(x, y) = x2 + xy + y2 + x− y + 1, x2 + y2 − 1 = 0;

9. f(x, y) =1

x+

1

y,

1

x2+

1

y2=

1

4, x, y 6= 0;

10. f(x, y) =x

3+y

4, x2 + y2 = 1;

11. f(x, y) = x2 − 3x+ (y − 1)2, x2 + y2 = 1.

26

Rezolvare:

1. Construim functia lui Lagrange L(x, y) = f(x, y) − αg(x, y) adica,L(x, y) = x2 + y2 − α(2x+ 3y − 1).

Determinam solutia sistemului

L′x(x, y) = 0

L′y(x, y) = 0

g(x, y) = 0

⇐⇒

2x− 2α = 0

2y − 3α = 0

2x+ 3y − 1 = 0

⇐⇒

x =2

13, y =

3

13, α =

2

13.

L′′x2(x, y) = L′′y2(x, y) = 2, L′′xy(x, y) = 0.

Deoarece d2L

(2

13,

3

13

)= 2(dx2 + dy2) este pozitiv definita, rezulta ca(

2

13,

3

13

)este punct de minim conditionat.

2. L(x, y) = xy − α(−x+ y − 1), L′x = y + α, L′y = x− α.

Solutia sistemului

L′x(x, y) = 0

L′y(x, y) = 0

g(x, y) = 0

este x = −1

2, y =

1

2, α = −1

2.

Deoarece d2L

(−1

2,1

2

)= 2dxdy diferentiem legatura ın punctul

(−1

2,1

2

)si obtinem −dx + dy = 0, dx = dy, deci d2L

(−1

2,1

2

)= 2dx2 > 0. Prin

urmare punctul

(−1

2,1

2

)este punct de minim conditionat.

34. Sa se determine punctele de extrem ale urmatoarelor functii f : R3 →R, cu legaturile specificate:

1. f(x, y, z) = x+ y + z, x+ y − z = 2, x2 + y2 + z2 = 4;

2. f(x, y, z) = x2 + y2 − z2, x+ y + z = 1, x2 + y2 + z2 = 4;

3. f(x, y, z) = x3 + y3 + z3, x2 + y2 + z2 = 3, x, y, z > 0;

4. f(x, y, z) = x+ 2y − 2z, x2 + y2 + z2 = 9;

5. f(x, y, z) = 2x− 3y + z, x2 + y2 + z2 = 14;

6. f(x, y, z) = x+ 2y − 2z, x2 + y2 + z2 = 16;

7. f(x, y, z) = xyz, x+ y + z = 5, xy + yz + zx = 8;

8. f(x, y, z) = x+ y + z,1

x+

1

y+

1

z= 1, x, y, z ∈ R∗+;

27

9. f(x, y, z) = x2 + y2 + z2,x

a+y

b+z

c= 1, a, b, c > 0.

Rezolvare:

1. Construim functia lui Lagrange L(x, y, z) = f(x, y, z)− αg1(x, y, z)−βg2(x, y, z), adica L(x, y, z) = x+y+z−α(x+y−z−2)−β(x2+y2+z2−4).

Determinam solutiile sistemului

L′x(x, y, z) = 0

L′y(x, y, z) = 0

L′z(x, y, z) = 0

g1(x, y, z) = 0

g2(x, y, z) = 0

temului

1− α− 2βx = 0

1− α− 2βy = 0

1 + α− 2βz = 0

x+ y − z − 2 = 0

x2 + y2 + z2 − 4 = 0

, care sunt a1 = (0, 0,−2), α1 = 1, β1 =

−1

2si a2 =

(4

3,4

3,2

3

), α2 = −1

3, β2 =

1

2.

Deoarece L′′x2(x, y, z) = L′′y2(x, y, z) = L′′z2(x, y, z) = −2β, L′′xy(x, y, z) =

L′′xz(x, y, z) = L′′yz(x, y, z) = 0 si d2L(x, y, z) = −2β(dx2+dy2+dz2), obtinemd2L(a1) = dx2 + dy2 + dz2 > 0, d2L(a2) = −(dx2 + dy2 + dz2) < 0, deci a1este punct de minim conditionat, iar a2 este punct de maxim conditionat.

2. Fie L(x, y, z) = x2 + y2 − z2 − α(x+ y + z − 1)− β(x2 + y2 + z2 − 4).

L′x(x, y, z) = 0

L′y(x, y, z) = 0

L′z(x, y, z) = 0

g1(x, y, z) = 0

g2(x, y, z) = 0

2x− α− 2βx = 0

2y − α− 2βy = 0

−2z − α− 2βz = 0

x+ y + z − 1 = 0

x2 + y2 + z2 − 4 = 0

a1 =

(2 +√

22

6,2 +√

22

6,4−√

22

6

), β1 =

2 +√

22√22− 1

respectiv

a2 =

(2−√

22

6,2−√

22

6,4 +√

22

6

), β2 = − 3

1 +√

22,

d2L(x, y, z) = 2(1− β)dx2 + 2(1− β)dy2 − 2(1 + β)dz2.Diferentiind legaturile obtinem dx+dy+dz = 0, xdx+ ydy+ zdz = 0 de

unde rezulta d2L(a1) =4(√

22− 4)√22− 1

dx2 > 0, d2L(a2) =4(√

22 + 4)√22 + 1

dx2 > 0,

adica a1 si a2 sunt puncte de minim conditionat.

28

0.5 Formula lui Taylor pentru functii de o

variabila reala

35. Sa se scrie formula lui Taylor cu restul sub forma lui Lagrange ın punctula = 0, pentru urmatoarele functii:

1. f(x) = ex;

2. f(x) = e−x;

3. f(x) =1

1 + x;

4. f(x) = ln(1− x2);

5. f(x) = ln

(1 + x

1− x

);

6. f(x) = arctan x;

7. f(x) = sinx;

8. f(x) = cos x.

Rezolvare:

1. f (n)(x) = ex, ∀n ∈ N, ∀x ∈ R, f (n)(0) = 1, ∀n ∈ N.Aplicand formula lui Taylor cu restul sub forma Lagrange obtinem

ex = 1 +x

1!+x2

2!+ · · ·+ xn

n!+

xn+1eξ

(n+ 1)!, ∀n ∈ N, unde ξ este situat ıntre

0 si x.

3. f (n)(x) = (−1)nn!(1 + x)−(n+1), ∀n ∈ N, ∀x ∈ R− {−1}.Aplicand formula lui Taylor cu restul sub forma Lagrange, obtinem

1

1 + x= 1−x+x2−x3+· · ·+(−1)nxn+(−1)n+1xn+1(1+ξ)−(n+2), ∀n ∈ N,

unde ξ este situat ıntre 0 si x.

36. Sa se dezvolte polinomul f(x) = x3 − 2x2 + 3x + 5 dupa puterileıntregi ale binomului x− 1.

Rezolvare:

Deoarece

f ′(x) = 3x2 − 4x+ 3, f ′′(x) = 6x− 4, f ′′′(x) = 6, f (n)(x) = 0, ∀n ≥ 4,

f(1) = 7, f ′(1) = 2, f ′′(1) = 2, f ′′′(1) = 6,

29

conform formulei lui Taylor, avem f(x) = 7 + 2(x− 1) + (x− 1)2 + (x− 1)3.

37. Sa se dezvolte polinomul f(x) = 2x4 − 3x2 + x − 1 dupa puterileıntregi ale binomului x+ 1.

38. Sa se dezvolte polinomul f(x) = x5 +x3 +x+ 1 dupa puterile ıntregiale binomului x− 2.

39. Sa se evalueze eroarea comisa ın aproximarea e ' 1 +1

1!+

1

2!+

1

3!.

40. Sa se calculeze valoarea aproximativa si apoi eroarea comisa, pentru:

1.√e;

2. 3√

12;

3.√

143;

4. 3√e;

5. ln(0, 9) .

Rezolvare:

1. Fie f(x) = ex. Aproximam functia f(x) cu polinomul Taylor de ordinul

n = 3, ın punctul a = 0, pentru x =1

2.

Intrucat f (n)(x) = ex,∀n ∈ N,∀x ∈ R obtinem

f

(1

2

)' T3;0

(1

2

)= 1 +

1

2+

1

8+

1

48=

79

48= 1, 646.

Eroarea comisa ın aceasta aproximare este E =

∣∣∣∣f (1

2

)− T3;0

(1

2

)∣∣∣∣ =

4!24, unde ξ ∈

(0,

1

2

). Prin urmare E <

2

4!24=

1

4!23=

1

192.

41. Sa se evalueze eroarea comisa ın aproximarile urmatoare:

1. cosx ' 1− x2

2+x4

24, x ∈ [0, 1];

2. sinx ' x− x3

6, x ∈ [0,

1

2];

3. 3√

1 + x ' 1 +x

3− x2

9, x ∈ [0, 1];

4. ex ' 1 + x+x2

2+x3

6, x ∈ [−1, 1];

30

5. e−x ' 1− x+x2

2− x3

6, x ∈ [0, 1];

6. coshx ' 1 +x2

2, x ∈ [0, 1];

7. sinhx ' x+x3

6, x ∈ [0, 1];

8.√

1 + x ' 1 +x

2− x2

8, x ∈ [0, 1] .

Rezolvare:

1. Eroarea comisa ın aceasta aproximare este E = |f(x)−Tn,a(x)| pentru

n = 4 si a = 0. Atunci E =

∣∣∣∣x5f (5)(ξ)

5!

∣∣∣∣ =

∣∣∣∣∣∣x5 cos

(ξ +

2

)5!

∣∣∣∣∣∣ ≤ 1

5!.

42. Sa se calculeze cu ajutorul formulei lui Taylor urmatoarele limite:

1. limx→0

√1 + x3 − 1

x3;

2. limx→0

ln(1 + 2x)− sin 2x+ 2x2

x3;

3. limx→0

3√

1 + x2 − 1

x2;

4. limx→1

3√x+√x− 2

x− 1;

5. limx→0

cos 7x− cos 3x

x2;

6. limx→0

√cos 3x− 1

x2;

7. limx→∞

[x− x2 ln

(1 +

1

x

)];

8. limx→0

cosh 2x− 2

x2;

9. limx→0

ex sinx− x− x2

x3;

31

10. limx→0

tanx− sinx

x3.

Rezolvare:

1. Folosim formula lui Taylor pentru functia f(x) =√

1 + x3 − 1, ınpunctul a = 0 cu restul sub forma Lagrange de ordin n = 3.

Avem

f ′(x) =3x2

2√

1 + x3, f ′′(x) =

3(4x+ x4)

4(1 + x3)√

1 + x3,

f ′′′(x) =3(8− 20x3 − x6)

8(1 + x3)2√

1 + x3, f(0) = f ′(0) = f ′′(0) = 0, f ′′′(0) = 3.

Cu ajutorul formulei lui Taylor obtinem f(x) =x3

2+x4f (4)(ξ)

4!.

Prin urmare limx→0

f(x)

x3= lim

x→0

(1

2+xf (4)(ξ)

4!

)=

1

2.

7. limx→∞

[x− x2 ln

(1 +

1

x

)]= lim

x→∞

1

x− ln

(1 +

1

x

)1

x2

= limt→0

t− ln(1 + t)

t2.

Aplicam formula lui Taylor pentru functia f(t) = t− ln(1 + t) ın punctula = 0 cu restul sub forma Lagrange de ordin n = 2.

Avem f ′(t) =t

1 + t, f ′′(t) =

1

(1 + t)2, f(0) = 0, f ′(0) = 0, f ′′(0) = 1.

Obtinem limt→0

f(t)

t2= lim

t→0

(1

2+tf (3)(ξ)

3!

)=

1

2.

43. Sa se determine numarul natural n, astfel ıncat ın aproximareaf(x) ' Tn,a(x), pentru functia f(x) =

√1 + x si a=0, eroarea comisa sa

fie cel mult1

16pe intervalul [0,1].

Rezolvare:

f ′(x) =1

2(1 + x)−

12 , f ′′(x) = − 1

22(1 + x)−

32 , · · · ,

f (n)(x) = (−1)n−11 · 3 · 5 · · · (2n− 3)

2n(1 + x)−

2n−12 , ∀n ≥ 2.

Trebuie sa determinam numarul natural n astfel ıncat |f(x)− Tn,0(x)| ≤1

16.

Avem

|f(x)−Tn,0(x)| =∣∣∣∣xn+1f (n+1)(ξ)

(n+ 1)!

∣∣∣∣ =1 · 3 · 5 · · · (2n− 1)

(n+ 1)!2n+1(1+ξ)−

2n+12 xn+1 ≤

1 · 3 · 5 · · · (2n− 1)

(n+ 1)!2n+1≤ 1

16de unde rezulta n ≥ 2.

32

44. Sa se determine cel mai mic numar natural n, astfel ıncat ın aproxi-marea f(x) ' Tn,a(x), pentru functia f(x) = ln(1+x) si a=0, eroarea comisasa fie cel mult 10−3 pe intervalul [0,1].

0.6 Formula lui Taylor pentru functii de mai

multe variabile reale

45. Sa se scrie formula lui Taylor pentru urmatoarele functii ın puncteleindicate:

1. f(x, y) = ex+y, a = (−1, 1);

2. f(x, y) = ln(1 + x+ y), a = (0, 0), n = 3;

3. f(x, y) = ln(1 + x)(1 + y), a = (0, 0), n = 3;

4. f(x, y) = ex cos y, a = (0, 0), n = 3;

5. f(x, y) = −x2 + 3xy2 − 15x+ 12y + 2, a = (1, 1);

6. f(x, y) = x2 + y2 + xy − 2x− 2y + 6, a = (1, 1);

7. f(x, y) = ex sin y, a = (0, 0), n = 3.

Rezolvare:

1. Deoarece f(n)

xkyn−k(x, y) = ex+y, ∀k = 0, n, atunci conform formuleilui Taylor pentru functii de mai multe variabile, rezulta ca exista punc-tul (ξ, η) situat ıntr-o vecinatate a punctului a astfel ıncat ∀(x, y) ın aceea

vecinatate sa avem f(x, y) = 1 +1

1!(x+ y) +

1

2!(x+ y)2 + · · ·+ 1

n!(x+ y)n +

1

(n+ 1)!dn+1f(ξ, η)(x+ 1, y − 1).

2. f(x, y) =1

1!(x+ y)− 1

2(x+ y)2 +

1

3(x+ y)3 +

1

4!d4f(ξ, η)(x, y).

46. Folosind formula lui Taylor de ordinul al doilea, sa se calculeze va-loarea aproximativa pentru:

1.√

0, 98 3√

1, 01;

2.√

4, 02 3√

7, 96;

3. (0, 95)1,01;

33

4. (0, 99)2(3, 01)3;

5. 1, 02(2, 02)2(3, 02)3 .

Rezolvare:

1. Pentru a determina valoarea aproximativa a√

0, 98 3√

1, 01 consideramfunctia f(x, y) =

√x 3√y si vom aproxima f(0, 98; 1, 01) cu polinomul Taylor

de ordinul al doilea ın punctul (1, 1), adica cu T2;(1,1)(0, 98; 1, 01).

Atunci f(x, y) ' 1 +1

1!

(1

2(x− 1) +

1

3(y − 1)

)+

+1

2!

(−1

4(x− 1)2 − 1

3(x− 1)(y − 1)− 2

9(y − 1)2

).

Punand x = 0, 98 si y = 1, 01 obtinem√

0, 98 3√

1, 01 ' 1 +1

1!

(−1

20, 02 +

1

30, 01

)+

+1

2!

(−1

4(0, 02)2 − 1

3(0, 02)(0, 01)− 2

9(0, 01)2

)=

89383

9000' 0, 994.

47. Sa se verifice daca sunt adevarate urmatoarele aproximari ın vecinatateapunctului (0, 0):

1. ln(1 + x) ln(1 + y) ' xy;

2. (1 + x)m(1 + y)n ' 1 +mx+ ny, m, n ∈ N∗;

3.cosx

cos y' 1− x2 − y2

2;

4. arctanx+ y

1 + xy' x+ y.

Rezolvare:

1. Fie f(x, y) = ln(1 + x) ln(1 + y). Atunci f(x, y) ' T2;(0,0)(x, y) esteechivalent cu ln(1 + x) ln(1 + y) ' xy.

0.7 Indicatii si raspunsuri

1. 3. f ′x(1,−2) = 3e+e−5, f ′y(1,−2) = e+3e−5; 4. f ′x(1,−1) =5

3, f ′y(1,−1) =

−2

3; 5. f ′x(2, 1) =

1

5, f ′y(2, 1) =

2

5; 6. f ′x(2, 1) =

1

2, f ′y(2, 1) = 0.

34

2. 3. |f(x, y)| ≤ |xy|, f continua ın (0,0), f ′x(0, 0) = f ′y(0, 0) = 0, f

diferentiabila ın (0,0); 4.

∣∣∣∣(x2 + y2) cos1

x2 + y2

∣∣∣∣ ≤ x2 + y2, f continua ın(0,0),

f ′x(0, 0) = f ′y(0, 0) = 0,

∣∣∣∣∣ x2 + y2√x2 + y2

cos1

x2 + y2

∣∣∣∣∣ ≤ √x2 + y2, f diferentiabila

ın (0,0).

3. 1. f ′x =x√

x2 + y2, f ′y =

y√x2 + y2

, df(x, y) =1√

x2 + y2(xdx +

ydy); 2. f ′x = 4xy − 3y + 2x, f ′y = 2x2 − 3x − 1, df(x, y) = (4xy − 3y +

2x)dx+ (2x2 − 3x− 1)dy; 3. f ′x =1

x, f ′y =

1

y; 4. f ′x = ln y, f ′y =

x

y; 5. f ′x =

−(lnx+1), f ′y = −(ln y+1); 6. f ′x =1

x, f ′y = −1

y; 7. f ′x = ln y−ey2+x, f ′y =

x

y− 2yey

2+x; 8. f ′x = (1 + xy)exy, f ′y = x2exy; 9. f ′x = 2xex2−y, f ′y =

−ex2−y; 10. f ′x = yxy−1, f ′y = xy lnx; 11. f ′x = (2x sin2 x+sin 2x)ex2+y2 ; f ′y =

2yex2+y2 sin2 x; 12. f ′x =

y

x2 + y2, f ′y = − x

x2 + y2;

13. f ′x =x

(1 + x2 + y2)√x2 + y2

, f ′y =y

(1 + x2 + y2)√x2 + y2

; 14. f ′x =

2x arctan(x2 + y2) +2x3

1 + (x2 + y2)2, f ′y =

2x2y

1 + (x2 + y2)2;

15. f ′x = − y

x√x2 − y2

, f ′y =1√

x2 − y2; 16. f ′x = − x√

1− x2 − y2,

f ′y = − y√1− x2 − y2

; 17. f ′x =1√

x2 + y2, f ′y =

y

(x+√x2 + y2)

√x2 + y2

;

18. f ′x =4xy2

(x2 + y2)2, f ′y = − 4x2y

(x2 + y2)2; 19. f ′x = 2x−3y+3, f ′y = −3x+4y−

4; 20. f ′x = 2 cos(2x+3y), f ′y = 3 cos(2x+3y); 21. f ′x = 3x2 +3y, f ′y = 3y2 +3x; 22. f ′x = (2x−a−b)(y−a)(y−b), f ′y = (2y−a−b)(x−a)(x−b); 23. f ′x =

(1+x)y2ex−y, f ′y = xy(2−y)ex−y; 24. f ′x = 2x3y3(2−3x2+2y), f ′y = x4y2(3−3x2 + 4y); 25. f ′x = − sinx+ cos(x+ y), f ′y = − sin y + cos(x+ y); 26. f ′x =

sin y sin(2x+y), f ′y = sinx sin(x+2y); 27. f ′x =2x

3(x2 + y2)+

y

1 + x2y2, f ′y =

2y

3(x2 + y2)+

x

1 + x2y2; 28. f ′x =

2x

nm(x2 + y), f ′y =

1

nm(x2 + y);

29. f ′x = 2x siny

x− x

2y − y3

x2cos

y

x; f ′y = −2y sin

y

x+x2 − y2

xcos

y

x; 30. f ′x =

(2xy + y3) sin 2(x2y + xy3), f ′y = (x2 + 3xy2) sin 2(x2y + xy3); 31. f ′x =

35

x2 − y2

x2y, f ′y =

y2 − x2

xy2; 32. f ′x =

y3(−3x4 + y2)

(x2 + y2)2, f ′y =

xy2(3x4 + y2)

(x2 + y2)2;

33.f ′x =y2 − x2

(x2 + y2)2cos

x

x2 + y2, f ′y =

−2xy

(x2 + y2)2cos

x

x2 + y2;

34. f ′x = 2x cos1√

x2 + y2+

x√x2 + y2

sin1√

x2 + y2,

f ′y = 2y cos1√

x2 + y2+

y√x2 + y2

sin1√

x2 + y2; 35. f ′x =

x2 − yx2y

,

f ′y =−x− 1

y2.

4. 1. f ′x = sin yz, f ′y = sin yz + (x+ y)z cos yz, f ′z = y(x+ y) cos yz,df(x, y, z) = sin yzdx+(sin yz+(x+y)z cos yz)dy+y(x+y) cos yzdz; 2. f ′x =2x + yz, f ′y = 2y + xz, f ′z = xy, df(x, y, z) = (2x + yz)dx + (2y + xz)dy +

xydz; 3. f ′x = 6x+ 2y, f ′y = 4y + 2x− 3z, f ′z = 6z − 3y; 4. f ′x = −3x2y2z +

10y, f ′y = −2x3yz + 10x, f ′z = −x3y2 − 6z2; 5. f ′x =x√

x2 + y2 + z2, f ′y =

y√x2 + y2 + z2

, f ′z =z√

x2 + y2 + z2; 6. f ′x =

1√y2 + z2

,

f ′y = − xy

(y2 + z2)√y2 + z2

, f ′z = − xz

(y2 + z2)√y2 + z2

; 7. f ′x =yz

x2y2 + z2,

f ′y =xz

x2y2 + z2, f ′z = − xy

x2y2 + z2; 8. f ′x = − xz

(x2 + y2 + z2)√x2 + y2

, f ′y =

− yz

(x2 + y2 + z2)√x2 + y2

, f ′z =

√x2 + y2

x2 + y2 + z2; 9. f ′x = −1

x, f ′y = −1

y, f ′z =

1

z; 10. f ′x = ln z+

y

x, f ′y = lnx+

z

y, f ′z = ln y+

x

z; 11. f ′x = − x

x2 + z2, f ′y =

1

y, f ′z = − z

x2 + z2; 12. f ′x = zex−y

2

+ yex+z, f ′y = −2yzex−y2

+ ex+z, f ′z =

ex−y2

+ yex+z; 13. f ′x = yxy−1 + zx ln z, f ′y = zyz−1 + xy lnx, f ′z = xzx−1 +

yz ln y; 14. f ′x = (yz3 cos(x+y)−sin(x+y))exyz3

, f ′y = (xz3 cos(x+y)−sin(x+

y))exyz3

, f ′z = 3xyz2 cos(x+y)exyz3

; 15. f ′x =x2 − yzx2y

, f ′y = −xz + y2

y2z, f ′z =

xy + z2

xz2; 16. f ′x = (y + b)(z + c), f ′y = (x + a)(z + c), f ′z = (x + a)(y +

b); 17. f ′x = yzexy, f ′y = xzexy, f ′z = exy; 18. f ′x = 2xzex2+y2 + yez, f ′y =

2yzex2+y2 + xez, f ′z = ex

2+y2 + xyez; 19.f ′x = cos y sin(2x+ z),f ′y = − sinx sin y sin(x+ z), f ′z = sinx cos y cos(x+ z);

20. f ′x =−2x3 + x2y − yz2

x2e

yx , f ′y =

z2 − x2

xe

yx , f ′z = 2ze

yx ; 21. f ′x =

36

y2z3(6 − 4x + 3y − 4z), f ′y = xyz3(12 − 4x + 9y − 8z), f ′z = xy2z2(18 −

6x+ 9y− 16z); 22. f ′x =z(x4 + 3x2z2 + x2y2 − y2z2)

(x2 + z2)2, f ′y = − 2xyz

x2 + y2, f ′z =

x(x2 − z2)(x2 − y2)(x2 + z2)2

; 23. f ′x =n(x− y)n−1

(y − z)m,

f ′y =(x− y)n−1(−mx+ (m− n)y + nz)

(y − z)m+1, f ′z =

m(x− y)n

(y − z)m+1; 24. f ′x = −y

2

x2,

f ′y =2y3 − xzxy2

, f ′z =2y + z2

yz2; 25. f ′x = −a sin 2(ax+ by + cz),

f ′y = −b sin 2(ax+ by + cz), f ′z = −c sin 2(ax+ by + cz).

5. 3. d2f(3, 1) = −(

2

9dx2 + dy2

); 4. d2f(1, 2) = −dx2 +

1

2dy2;

5. d2f(−1, 1) =2

25(−3dx2 + 16dxdy − 3dy2); 6. d2f(0, 0) = m(m− 1)dx2 +

2mndxdy + n(n− 1)dy2; 7. d2f(0, π) = −dx2 + dy2; 8. d2f(2, 1) = 4(dx2 −3dxdy+2dy2); 9. d2f(1, 1) = 4(dx2+dy2); 10. d2f(1,−1) = 2(3dx2+4dxdy+dy2).

6. 2. d2f(1, 1, 1) =2

3√

3(dxdy + dydz + dxdz); 3. d2f(1, 1, 1) = 2e(dx2 +

dy2 + dz2 + 4dxdy + 4dydz + 4dxdz); 4. d2f(1, 2,−1) = 4(39dx2 + 3dy2 +5dz2 + 27dxdy − 24dxdz); 5. d2f(1, 1, 1) = e(dx2 + dy2 + 3dz2 + 4dxdy +

6dydz+4dxdz); 6. d2f(2, 1, 1) = −dy2+dz2+dxdy−dxdz; 7. d2f(π,π

2,π

3) =

dx2 +dy2 + dz2− 2dxdy+ 2dydz− 2dxdz; 8. d2f(1,−1, 2) = 2(4dx2 + 3dy2 +

2dz2−10dxdy+5dydz−6dxdz); 9. d2f(2, 1, 1) = − 1

3√

3(7dx2+4dy2+4dz2+

4dxdy + 2dydz + 4dxdz); 10. d2f(2, 0, 1) =1

9(2dx2 − 4dy2 − 4dz2 − 2dxdy +

10dydz − 2dxdz).

7. 2. f ′′x2 = 12x2 − 4, f ′′xy = 5, f ′′y2 = 12y2 − 90y4, d2f(x, y) = (12x2 −4)dx2 + 10dxdy + (12y2 − 90y4)dy2;

3. d2f(x, y) =14

(x+ 3y)3(−ydx2 + (x− 3y)dxdy + 3xdy2);

4. d2f(x, y) = (dx2 − 4ydxdy + (4y2 − 2)dy2)ex−y2

; 5. d2f(x, y) = (y3dx2 +2(xy2 + 2y)dxdy + (x2y + 2x)dy2)exy; 6.d2f(x, y) = ((x2 + 4x + y2 + 2)dx2 + 2(x2 + y2 + 2x + 2y)dxdy + (x2 +

y2 + 4y+ 2)dy2)ex+y; 7. d2f(x, y) =1

(x2 + y3)2(2(y3−x2)dx2− 12xy2dxdy+

3y(2x2− y3)dy2); 8. d2f(x, y) = −y2 sinxydx2 + 2(cosxy− xy sinxy)dxdy−x2 sinxydy2; 9. d2f(x, y) =

2

(x2 + y2)(−xydx2 + (x2 − y2)dxdy + xydy2);

37

10. d2f(x, y) = yx−2(y2 ln2 ydx2 + 2y(x ln y + 1)dxdy + x(x− 1)dy2).

8. 2. d2f(x, y, z) =2

(y + z)3((x + z)dy2 + (x − y)dz2 − (y + z)dxdy +

(2x− y+ z)dydz− (y+ z)dxdz); 3. d2f(x, y, z) = 2(−y2z3dx2 + xz3(6− x+6y − z)dy2 + xy2z(18 − 3x + 6y − 6z)dz2 + yz3(12 − 4x + 6y − 2z)dxdy +xyz2(36−6x+18y−8z)dydz+y2z2(18−6x+6y−4z)dxdz); 4. d2f(x, y, z) =zexdx2−xeydy2+(2+yez)dz2−2eydxdy+2ezdydz+2exdxdz; 5. d2f(x, y, z) =2xyz + (x+ z)y2 − 2x2y

x4e

yxdx2 +

x+ z

x2e

yxdy2 − 2(xz + xy + yz)

x3e

yxdxdy +

2

xe

yxdydz − 2y

x2e

yxdxdz;

6. d2f(x, y, z) =1

4(x2 + z)32

(4yzdx2−ydz2+8x(x2+z)dxdy+4(x2+z)dydz−

4xydxdz); 7. d2f(x, y, z) = − 1

(x+ y + z)2(dx + dy + dz)2; 8. d2f(x, y, z) =

−(

1

x2dx2 +

1

y2dy2 +

1

z2dz2)

; 9. d2f(x, y, z) = − cos(x+2y+3z)(dx+2dy+

3dz)2; 10. d2f(x, y, z) = −z sinxdx2− x sin ydy2− y sin zdz2 + 2 cos ydxdy+2 cos zdydz + 2 cosxdxdz.

9. 2. Jf (r, θ, ϕ) =

cos θ cosϕ −r sin θ cosϕ −r cos θ sinϕsin θ cosϕ r cos θ cosϕ −r sin θ sinϕ

sinϕ 0 r cosϕ

,

detJf (r, θ, ϕ) = r2 cosϕ; 3. Jf (1, 1) =

(2 −14 1

), detJf (1, 1) = 6.

10. 1. Jf (x, y) =

y x1 11 −1

; 2. Jf (x, y, z) =

yz xz xy1

yz− x

y2z− x

yz2

;

3. Jf (x, y, z) =

− a

x2z y

b − 1

y21

z2

.

11. 3. gradf (x, y) =

(2x

x2 + y2, ln(x2 + y2) +

2y2

x2 + y2

(−3,−1); 5. gradf (x, y, z) =

(− yz

x2y2 + z2,− xz

x2y2 + z2,

xy

x2y2 + z2

);

6. gradf (x, y, z) = (a cos(ax + by + cz), b cos(ax + by + cz), c cos(ax + by +cz)); 7. gradf (−1, 1, 2) = (−2, 2, 4).

12. 2.df

du(1, 1) = −1; 3.

df

du(−1, 1) =

18√

10

5; 4.

df

du(x, y) = 1;

38

5.df

du(1, 1) = −

√5

10; 6.

df

du(1, 2,−1) = −2e

3− 1

6(e2 − e−2); 7.

df

du(1,−1, 1) =

2√

3;

8.df

du(1,−1, 2) = 15; 9.

df

du(1, 0,

π

4) = 0.

13. 2. f ′(x) = nun−1vm cosx−munvm−1 sinx; 4. f ′(x) =2v lnx

x+

6ux tan2(x2 + 1)

cos2(x2 + 1); 5. f ′(x) =

xvex − uxv2

; 6. f ′(x) = (v cosx−u lnu sinx)uv−1;

7. f ′(x) =1√v

tanu√v

(−6x+

ux

2v√x2 + 1

).

14. 3. f ′x =1

u+y

v, f ′y =

1

u+x

v; 4. f ′x = (nun−1−nmunm−1vnm) cos(x−y)−

y(mvm−1−nmunmvnm−1) sinxy, f ′y = −(nun−1−nmunm−1vnm) cos(x− y)−

x(mvm−1 − nmunmvnm−1) sinxy; 5. f ′x =

(2uv2

x+ y+ y(1 + u2v)xy−1

)eu

2v,

f ′y =

(2uv2

x+ y+ (1 + u2v)xy lnx

)eu

2v; 6. f ′x = 0, f ′y = 1.

15. 3. u = x3y−5x2+2xy2+4x+1, v = x2y5−7y, f ′x = (3x2y−10x+2y2+

4)ϕ′u + 2xy5ϕ′v, f′y = (x3 + 4xy)ϕ′u + (5x2y4 − 7)ϕ′v; 4. u = xy, v =

y

x, f ′x =

yϕ′u −y

x2ϕ′v, f

′y = xϕ′u +

1

xϕ′v; 5. u = xyz, v = x2 − y2 + z2, f ′x = yzϕ′u +

2xϕ′v, f′y = xzϕ′u− 2yϕ′v, f

′z = xyϕ′u + 2zϕ′v; 6. u = x+ yz, v = ex

2y+z, w =

lnxyz, f ′x = ϕ′u + 2x2yex2y+zϕ′v +

1

xϕ′w, f

′y = zϕ′u + x2ex

2y+zϕ′v +z

yϕ′w, f

′z =

yϕ′u + ex2y+zϕ′v + ln yϕ′w; 7. u =

√x2 + y2 + z2, v = x arctan

y

z, f ′x =

x√x2 + y2 + z2

ϕ′u + arctany

zϕ′v, f

′y =

y√x2 + y2 + z2

ϕ′u +xz

y2 + z2ϕ′v, f

′z =

z√x2 + y2 + z2

ϕ′u −xy

y2 + z2ϕ′v; 8. u = e−x

2+yz, v = xeyz, w = exyz, f ′x =

−2xe−x2+yzϕ′u+e

yzϕ′v+yzexyzϕ′w, f

′y = ze−x

2+yzϕ′u+xzeyzϕ′v+xze

xyzϕ′w, f′z =

ye−x2+yzϕ′u + xyeyzϕ′v + xyexyzϕ′w.

16. 3. u = lnxy, v = exy, d2f(x, y) =

=

(− 1

x2ϕ′u + y2exyϕ′v +

1

x2ϕ′′u2 +

2y

xexyϕ′′uv + y2e2xyϕ′′v2

)dx2 +

+ 2

(1

xyϕ′′u2 + 2exyϕ′′uv + (1 + xy)exyϕ′v + xye2xyϕ′′v2

)dxdy +

+

(− 1

y2ϕ′u + x2exyϕ′v +

1

y2ϕ′′u2 +

2x

yexyϕ′′uv + x2e2xyϕ′′v2

)dy2; 4. u = x+ y −

39

z, v = x− y+ z, w = −x+ y+ z, d2f(x, y, z) = (ϕ′′u2 + 2ϕ′′uv − 2ϕ′′uw +ϕ′′v2 −2ϕ′′vw +ϕ′′w2)dx2 + (ϕ′′u2 − 2ϕ′′uv + 2ϕ′′uw +ϕ′′v2 − 2ϕ′′vw +ϕ′′w2)dy2 + (ϕ′′u2 − 2ϕ′′uv−2ϕ′′uw + ϕ′′v2 + 2ϕ′′vw + ϕ′′w2)dz2 + 2(ϕ′′u2 − ϕ′′v2 − ϕ′′w2 + 2ϕ′′vw)dxdy + 2(−ϕ′′u2 −ϕ′′v2 + ϕ′′w2 + 2ϕ′′uv)dydz + 2(−ϕ′′u2 + ϕ′′v2 − ϕ′′w2 + 2ϕ′′uw)dxdz; 5. u = xyz, v =x

yz, d2f(x, y, z) =

(y2z2ϕ′′u2 + 2ϕ′′uv +

1

y2z2ϕ′′v2

)dx2 +

+

(x2z2ϕ′′u2 −

2x2

y2ϕ′′uv +

2x

y3zϕ′v +

x2

y4z2ϕ′′v2

)dy2 +

+

(x2y2ϕ′′u2 −

2x2

z2ϕ′′uv +

x2

y2z4ϕ′′v2 +

2x

yz3ϕ′v

)dz2 +

+ 2

(zϕ′u −

1

y2zϕ′v + xyz2ϕ′′u2 −

x

y3z2ϕ′′v2

)dxdy +

+ 2

(xϕ′u +

x

y2z2ϕ′v + x2yzϕ′′u2 −

2x2

yzϕ′′uv +

x2

y3z3ϕ′′v2

)dydz +

+2

(yϕ′u −

1

yz2ϕ′v + xy2zϕ′′u2 −

x

y2z3ϕ′′v2

)dxdz; 6. u = x+yz, v = x2 +y2−

z2, d2f(x, y, z) = (ϕ′′u2 +4xϕ′′uv +4x2ϕ′′v2 +2ϕ′v)dx2 +(z2ϕ′′u2 +4yzϕ′′uv +2ϕ′v +

4y2ϕ′′v2)dy2 + (y2ϕ′′u2 − 4yzϕ′′uv + 4z2ϕ′′v2 − 2ϕ′v)dz

2 + 2(zϕ′′u2 + 2(y + xz)ϕ′′uv +4xyϕ′′v2)dxdy+2(ϕ′u+yzϕ′′u2−2(z2−y2)ϕ′′uv−4yzϕ′′v2)dydz+2(yϕ′′u2 +2(−z+xy)ϕ′′uv − 4xzϕ′′v2)dxdz.

18. 2. f(n)xn = an sin

(ax+ by +

2

), f

(n)yn = bn sin

(ax+ by +

2

),

f(n)

xkyn−k = akbn−k sin(ax+ by +

2

); 4. f

(n)xn = (x2 + 2nx+ n(n− 1))ex+y,

f(n)yn = x2ex+y, f

(n)

xkyn−k = (x2 + 2kx + k(k − 1))ex+y; 5. f(n)xn = (−1)n+1(n −

1)!(x+y)−n, f(n)yn = (−1)n+1(n−1)!(x+y)−n, f

(n)

xkyn−k = (−1)n+1(n−1)!(x+

y)−n; 6. f(n)xn = aneax+by+cz, f

(n)yn = bneax+by+cz, f

(n)zn = cneax+by+cz, f

(n)

xkypzr=

akbpcreax+by+cz, k+p+ r = n; 7. f(n)xn = an cos

(ax+ by + cz +

2

), f

(n)yn =

bn cos(ax+ by + cz +

2

), f

(n)zn = cn cos

(ax+ by + cz +

2

), f

(n)

xkypzr=

akbpcr cos(ax+ by + cz +

2

), k + p + r = n; 8. f

(n)xn = (−1)n+1(n −

1)!an(ax + by + cz)−n, f(n)yn = (−1)n+1(n − 1)!bn(ax + by + cz)−n, f

(n)zn =

(−1)n+1(n−1)!cn(ax+ by+ cz)−n, f(n)

xkypzr= (−1)n+1(n−1)!akbpzr(ax+ by+

cz)−n, k + p+ r = n.

19. 1. f(n)xn = (xy + n)yn+2exy, f

(n)yn = (x3y3 + 3nx2y2 + 3n(n − 1)xy +

n(n − 1)(n − 2))xn−2exy 2. f(3)xyz = (1 + 3xyz + x2y2z2)exyz, f

(3)

x2y = (2yz2 +

xy2z3)exyz, f(3)

x3 = y3z3exyz; 3. f(3)

xz2 = (x+ y)(2 + (x+ y)(y + z))exy+yz+zx,

f(3)

y3 = (x+z)3exy+yz+zx, f(3)xyz = (2x+2y+2z+(x+y)(y+z)(x+z))exy+yz+zx.

40

20. 2. (-2,3) punct de minim local; 3.

(−4

3,−2

3

)punct de minim local;

4. (1,1) punct de minim local; 5. (0,0) punct de maxim local, (−√

2,√

2),

(√

2,−√

2) puncte de minim local; 7.

(−5

4,1

2

)punct de minim local;

8. (-1,2) punct de minim local; 9.

(2π

3,2π

3

),

(5π

3,5π

3

)puncte de minim

local,(π

3,π

3

),

(4π

3,4π

3

)puncte de maxim local; 10.

(−1

3,−1

3

)punct de

minim local; 11. (3,-2) punct de minim local; 12.

(2

5,2

5

)punct de maxim

local.

21. 2. (1,−3, 4) punct de minim local; 3.

(1

2, 1, 1

)punct de minim

local; 4. (-1,1,0) punct de minim local; 5.

(2

3,−1

3,−1

2

)punct de minim

local; 6. (2,4,8) punct de minim local; 7. nu are puncte de extrem local; 8.

(24,-144,-1) punct de minim local; 9.

(1,−2,

3

2

)punct de minim local; 10.

nu are puncte de extrem; 11. nu are puncte de extrem.

22. 2. f ′(1) = −1

5, f ′′(1) = −28

53; 3. f ′(0) =

2

3, f ′′(0) =

16

27; 4. f ′(0) =

−1

2, f ′′(0) =

3

2; 5. f ′(1) = −1, f ′′(1) = −22

5.

23. 2. f ′x(1, 0) = 1, f ′y(1, 0) = 0; 3. f ′x(0,−1) = −3, f ′y(0,−1) = 4;

4. f ′x(1,−1) =1

3, f ′y(1,−1) =

1

3; 5. f ′x(2, 2) =

1

3, f ′y(2, 2) =

1

3; 6. f ′x(0, 0) =

1, f ′y(0, 0) = −1.

24. 3. f ′(x) =ex − yey + x

,

f ′′(x) =ex+2y − e2x+y + 2(x+ y − 1)ex+y + 2yey − 2xex + x2ex − y2ey + 2xy

(ey + x)3;

4. f ′(x) = −2x+ y

2y + x, f ′′(x) = −6(x2 + y2 + xy)

(2y + x)3.

25. 2. f ′x =yz − x2

z2 − xy, f ′y =

xz − y2

z2 − xy, f ′′x2 =

2xz(3xyz − y3 − z3 − x3)(z2 − xy)3

, f ′′xy =

xy4 + x4y + x3z2 + z5 + y3z2 − 2xyz3 − 3x2y2z

(z2 − xy)3,

f ′′y2 =2yz(3xyz − x3 − z3 − y3)

(z2 − xy)3; 3. f ′x(x, y) = −x

z, f ′y(x, y) = −y

z, f ′′x2 =

41

−x2 + z2

z3, f ′′y2 = −y

2 + z2

z3, f ′′xy = −xy

z3.

26. 3. f ′(1) = −4

5, g′(1) =

1

5, f ′′(1) = −36

25, g′′(1) =

4

25; 4. f ′(x) =

−x− 2xz

y + z, g′(x) =

2xy − xy + z

;

5. f ′(x) = −xy, g′(x) = 0, f ′′(x) = −x

2 + y2

y3, g′′(x) = 0.

29. 3. x = 1 punct de maxim local, x =1

2punct de minim local; 4. x = 0

punct de minim local, x = −2 punct de maxim local; 5. x =5

8punct de

maxim local; 6.

(a

√3

8, a

√1

8

),

(−a√

3

8, a

√1

8

)puncte de maxim local,(

−a√

3

8,−a

√1

8

),

(a

√3

8,−a

√1

8

)puncte de minim local.

30. 2. (-1,2,1) punct de minim local, (-1,2,-2) punct de maxim local;3. (−3 −

√6,−3 −

√6,−4 − 2

√6) punct de minim local, (−3 +

√6,−3 +√

6,−4 + 2√

6) punct de maxim local.

31. 2. df(1, 1) = −dx− 2dy, dg(1, 1) = 0. d2f(1, 1) = −4dx2 − 8dxdy −10dy2, d2g(1, 1) = 8dxdy + 2dy2; 3. df(0, 1) = dx+ dy, dg(0, 1) = −dx,d2f(0, 1) = 0, d2g(0, 1) = 2dx2 + 2dxdy;

32. 2. f ′x =y

y + 1, f ′y =

v

y + 1, g′x =

1

y + 1, g′y = − v

y + 1; 3. f ′x(x, y) =

x2 − v2

v2 − u2, f ′y(x, y) =

y2 − v2

v2 − u2, g′x(x, y) =

u2 − x2

v2 − u2, g′y(x, y) =

u2 − y2

v2 − u2;

4. f ′x(x, y) =x cos v + sin v

x cos v + y cosu, f ′y(x, y) =

x cos v − sinu

x cos v + y cosu,

g′x(x, y) =y cosu− sin v

x cos v + y cosu, g′y(x, y) =

y cosu+ sinu

x cos v + y cosu.

33. 3.

(ab2

a2 + b2,

a2b

a2 + b2

)punct de minim conditionat; 4. (-2,-1) punct

de minim conditionat, (2,1) punct de maxim conditionat; 5.

(3

2,1

2

)punct

de minim conditionat,

(−3

2,−1

2

)punct de maxim conditionat; 6.

(1

2,1

2

)punct de maxim conditionat; 7. (0,1) punct de minim conditionat;

8.

(√2

2,−√

2

2

),

(−√

2

2,

√2

2

)puncte de minim conditionat, (0,-1), (1,0)

42

puncte de maxim conditionat; 9. (−2√

2,−2√

2) punct de minim conditionat,

(2√

2, 2√

2) punct de maxim conditionat; 10.

(−4

5,−3

5

)punct de minim

conditionat,

(4

5,3

5

)punct de maxim conditionat; 11.

(− 3√

13,− 2√

13

)punct de maxim conditionat,

(3√13,

2√13

)punct de minim conditionat.

34. 3. (1,1,1) punct de minim conditionat; 4. (-1,-2,2) punct de minimconditionat, (1,2,-2) punct de maxim conditionat; 5. (-2,3,-1) punct de

minim conditionat, (2,-3,1) punct de maxim conditionat; 6.

(−4

3,−8

3,8

3

)punct de minim conditionat,

(4

3,8

3,−8

3

)punct de maxim conditionat; 7.(

4

3,4

3,7

3

),

(4

3,7

3,4

3

),

(7

3,4

3,4

3

)puncte de maxim conditionat, (2,2,1),

(2,1,2), (1,2,2) puncte de minim conditionat; 8. (3,3,3) punct de minimconditionat;

9.

(1

a

(1

a2+

1

b2+

1

c2

)−1,1

b

(1

a2+

1

b2+

1

c2

)−1,1

c

(1

a2+

1

b2+

1

c2

)−1)punct de minim conditionat

35. 2. e−x = 1− x1!

+x2

2!+· · ·+(−1)n

xn

n!+(−1)n+1 xn+1

(n+ 1)!e−ξ, unde ξ este

situat ıntre 0 si x; 4. ln(1−x2) = −2

(x2

2+x4

4+x6

6+ · · ·+ ((−1)n−1 + 1)xn

n

)+ (−1)n

((1 + ξ)−n−1 + (ξ − 1)−n−1)xn+1

n+ 1;

5. ln

(1 + x

1− x

)= 2

(x+

x3

3+x5

5+ · · ·+ ((−1)n−1 − 1)xn

n

)+

+ (−1)n((1 + ξ)−n−1 − (ξ − 1)−n−1)xn+1

n+ 1; 6. f(x) = arctan x implica x =

tan f ; f ′(x) =1

1 + x2=

1

1 + tan2 f= cos2 f = cos f sin

(f +

π

2

), f ′′(x) =

sin(2f+π)f ′(x) = sin 2(f +

π

2

)cos2 f, f (3)(x) = 2·3 sin 3

(f +

π

2

)cos3 f, ...,

f (n)(x) = (n− 1)!(1 + x2)−n2 sinn

(arctanx+

π

2

)arctanx =

x

1− x3

3+x5

5+ · · ·+ sin

2

xn

n+

xn+1

n+ 1(1 + ξ2)−

n+12 sin

[(n+ 1)

(arctan ξ +

π

2

)]; 7. sin x =

x

1!− x3

3!+x5

5!+

43

· · ·+ sinnπ

2

xn

n!+

xn+1

(n+ 1)!sin(ξ +

2

);

8. cosx = 1− x2

2!+x4

4!− x6

6!+ · · ·+ cos

2

xn

n!+

xn+1

(n+ 1)!cos(ξ +

2

).

37. f(x) = −3− (x+ 1) + 9(x+ 1)2 − 8(x+ 1)3 + 2(x+ 1)4.

38. f(x) = 43+93(x−2)+86(x−2)2 +41(x−2)3 +10(x−2)4 +(x−2)5.

39. E = |f(x)− T3,0(x)| = eξx4

4!<

e

4!<

1

8.

40. 2. f(x) = 3√x, x = 12, a = 8, 3

√12 ' T3;8(12) =

743

324= 2, 29, E =

80

81x−

113 <

80

81

1

211=

5

81 · 27; 3. f(x) =

√x, x = 143, a = 144,

√143 '

T3;144(143) = 11, 958; 4. f(x) = ex, x =1

3, a = 0, 3

√e ' T3;0

(1

3

)=

1, 395; 5. f(x) = lnx, x = 0, 9, a = 1, ln 0, 9 ' T3;1(0, 9) = −0, 106.

41. 2. E = |f(x) − T3;0(x)| =

∣∣∣∣x44!sin(ξ +

2

)∣∣∣∣ ≤ 1

4!; 3. E = |f(x) −

T2;0(x)| =

∣∣∣∣∣ 10x3

27 · 3! 3√

(1 + ξ)8

∣∣∣∣∣ ≤ 5

34; 4. E = |f(x) − T3;0(x)| =

∣∣∣∣x4eξ4!

∣∣∣∣ ≤e

4!; 5. E = |f(x) − T3;0(x)| =

∣∣∣∣x44!e−ξ∣∣∣∣ ≤ 1

4!; 6. E = |f(x) − T2;0(x)| =∣∣∣∣(eξ − e−ξ)x32 · 3!

∣∣∣∣ ≤ 1

12

(e− 1

e

); 7. E = |f(x) − T3;0(x)| =

∣∣∣∣(eξ − e−ξ)x42 · 4!

∣∣∣∣ ≤1

48

(e− 1

e

); 8. E = |f(x)− T2;0(x)| =

∣∣∣∣ 3x3

8 · 3!(x+ 1)−

52

∣∣∣∣ ≤ 1

16.

42. 2. f(x) = ln(1 + 2x) − sin 2x + 2x2, a = 0, n = 3, limx→0

f(x)

x3=

4; 3. f(x) =3√

1 + x2 − 1, a = 0, n = 2, limx→0

f(x)

x2=

1

3; 4. f(x) = 3

√x +

√x− 2, a = 1, n = 1, lim

x→1

f(x)

x− 1=

5

6; 5. f(x) = cos 7x− cos 3x, a = 0, n =

2, limx→0

f(x)

x2= −20; 6. f(x) =

√cos 3x − 1, a = 0, n = 2, lim

x→0

f(x)

x2=

−9

4; 8. f(x) = cosh 2x − 2, a = 0, n = 2, lim

x→0

f(x)

x2= 4; 9. f(x) =

ex sinx− x− x2, a = 0, n = 3, limx→0

f(x)

x3=

1

3; 10. f(x) = tan x− sinx, a =

0, n = 3, limx→0

f(x)

x3=

1

2.

44

44. E = |f(x) − Tn,0(x)| =

∣∣∣∣(1 + ξ)−(n+1)xn+1

n+ 1

∣∣∣∣ ≤ 1

n+ 1≤ 1

103, de unde

n+ 1 ≥ 103, adica nmin = 103 − 1.

45. 3. ln(1 + x)(1 + y) =1

1!(x + y) − 1

2!(x2 + y2) +

2

3!(x3 + y3) +

1

4!d4f(ξ, η)(x, y);

4. ex cos y = 1 +1

1!x+

1

2!(x2 − y2) +

1

3!(x3 − 3xy2) +

1

4!d4f(ξ, η)(x, y);

5. f(x, y) = 1 +1

1!(−14(x− 1) + 18(y− 1)) +

1

2!(−2(x− 1)2 + 12(x− 1)(y−

1)+6(y−1)2)+1

3!(18(x−1)(y−1)2); 6. f(x, y) = 5+

1

1!(x+y−2)+

1

2!(2(x−

1)2 + 2(x− 1)(y− 1) + 2(y− 1)2); 7. ex sin y =1

1!y+

1

2!2xy+

1

3!(3x2y− y3) +

1

4!d4f(ξ, η)(x, y).

46. 2. f(x, y) =√x 3√y, (x, y) = (4, 02; 7, 96), (x0, y0) = (4, 8), f(x, y) '

T2,(x0,y0) = 4, 003; 3. f(x, y) = xy, (x, y) = (0, 95; 1, 01), (x0, y0) = (1, 1),f(x, y) ' T2,(x0,y0) = 0, 95; 4. f(x, y) = x2y3, (x, y) = (0, 99; 3, 01),(x0, y0) = (1, 3), f(x, y) ' T2,(x0,y0) = 26, 7282; 5. f(x, y) = xy2z3,(x, y, z) = (1, 02; 2, 02; 3, 02), (x0, y0, z0) = (1, 2, 3), f(x, y, z) ' T2,(x0,y0,z0) =114, 6348.

47. 2. f(x, y) ' T1,(0,0); 3. f(x, y) ' T2,(0,0); 4. f(x, y) ' T1,(0,0).

45

Documents
Documents
Documents
News & Politics
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Sports
Documents
Documents
Documents
Documents
Documents
Documents
Documents