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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.
,
adica
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
, adica
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
, adica solutiile sis-
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
)∣∣∣∣ =
eξ
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 +
nπ
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
); 4. gradf (1,−1) =
(−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 +
nπ
2
), f
(n)yn = bn sin
(ax+ by +
nπ
2
),
f(n)
xkyn−k = akbn−k sin(ax+ by +
nπ
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 +
nπ
2
), f
(n)yn =
bn cos(ax+ by + cz +
nπ
2
), f
(n)zn = cn cos
(ax+ by + cz +
nπ
2
), f
(n)
xkypzr=
akbpcr cos(ax+ by + cz +
nπ
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
nπ
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 +
nπ
2
);
8. cosx = 1− x2
2!+x4
4!− x6
6!+ · · ·+ cos
nπ
2
xn
n!+
xn+1
(n+ 1)!cos(ξ +
nπ
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 +
nπ
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
