Higher_Mathematics_Part_2

If

x and

y

are independent variables, it is given by

d

2

z

=

∂ 2

f

dx

2

+

2

∂ 2 f

dxdy +

∂ 2 f

dy

2

.

(1.8)

∂x

2

∂x∂y

∂y 2

Differential of order n d n z

of a function

z = f (x, y)

at point

M (x, y)

is

called the differential of the order (n - 1) differential

d n−1 z , thus

d n z = d (d n−1 z)

.

(1.9)

If

x and

y

are independent variables(1.9), it is given by

n

∂n f

d n z = ∑ Cnk

dxk dyn−k ,

∂xk ∂yn−k

k =0

where

Cnk =

n!

.

k!(n − k)!

Symbolically,

n

n

∂

∂

d

z =

dx

+

dy

f .

∂x

∂y

Typical problems

Т.2

1. Let z be given by the formula z = x2 + xy − 2y . Compute zx,

zy and

z

at point M (0;1) when

= x2 ,

y = −1 .

Solution. To compute z(0;1) :

z(0;1) = 02 + 0 1 − 2 1 = −2 ,

begin by computing the new values of the variables,

x + x = 0 + 2 = 2 ,

y +

y = 1 − 1 = 0 ;

z(x +

x, y +

y) = z(2; 0) = 4 ,

z(x +

x, y) = z(2;1) = 4 ,

z(x, y +

y) = z(0; 0) = 0 .

Then

z(0;1) = 4 − (−2) = 6 ,

x z = 4 − (−2) = 6 ,

y z = 0 − (−2) = 2 .

2. Find the partial derivatives for the given functions z.

а)

z = x5 y2 + 2y3 − x − 4 ;

б) u = x sin(yz) + (x + y2 + z3 )5 ;

в) z = xln y ;

г) u = x tg

y

+ z 3 2 x− z .

x

21

Solution. From the definition,

а)

∂z

= 5x4 y2 − 1,

∂z

= 2yx5

+ 6y 2 ;

∂x

∂y

б)

∂u

= sin(yz) + 5(x + y2 + z3 )4 ,

∂x

∂u

= x cos(yz) z + 10y(x + y2 + z3 )4 ,

∂y

∂u

= x cos(yz) y + 15z2 (x + y2 + z3 )4 ;

∂z

в)

∂z

= (ln y) xln y−1 ,

∂z

=

xln y ln x

1

= xln y

ln x

;

∂x

∂y

y

y

∂u

y

1

y

3

x− z

г)

= tg

+ x

−

+ z

2

ln 2 ,

∂x

x

cos2 ( y / x)

x 2

∂u

= x

1

1

=

1

,

∂y

cos2 ( y / x) x

cos2 ( y / x)

∂u

= 3z2 2x− z − z3 2x− z ln 2 = 2x− z (3z2 − z3 ) ln 2 .

∂z

F ≡ x2 + y2 z2 + xyz ,

∂F

= 2x + yz ,

∂x

∂F

=

2yz 2 + xz ,

∂F

= 2zy 2 + xy .

∂y

∂z

Hence

∂z

=

−

2x + yz

,

∂z

= −

2yz 2 + xz

.

∂x

2zy 2 + xy

∂y

2zy

2 + xy

4. Find the partial derivatives for the given function

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

Solution. This function is composite:

z = f (u, v) , де u = x + 2y , v = x2 y .

By formulas (1.2) find:

∂z

=

∂f

+

∂f

2xy ,

∂z

=

∂f

2 +

∂f

x2 .

∂x

∂u

∂v

∂y

∂u

∂v

22

x

z = (2x + y) y .

Solution. Using logarithmic differentiation, we have

ln z =

x

ln(2x + y) ;

(ln z)′

= (

x

ln(2x + y))′

;

y

x

y

x

1

∂z

=

1

ln(2x + y) +

x

2

;

∂z

=

z

1

ln(2x + y) +

x 2

.

z ∂x y

y 2x + y

∂x

y

y 2x + y

Hence

x

∂z

1

2x

y

= (2x + y)

ln(2x + y) +

.

∂x

y

2x

+ y

Likewise find

∂z

.

∂y

(ln z)′

= (

x

ln(2x

+ y))′

;

1

∂z

= −

x

ln(2x + y) +

x

1

;

y

y

y

z ∂y

y 2

y 2x + y

∂z

x

x

y

y

=

(2x + y)

− ln(2x

+ y) +

.

∂y

y

2

2x + y

а) z = e x y3 ;

б) u = x arcsin y + z 4 .

Solution.

а) dz = (e x y3 )′ dx + (e x y3 )′ dy = ex y3dx + 3ex y2 dy ;

x

y

б) ∂u

= arcsin y ,

∂u =

x

,

∂u = 4z3

;

∂x

∂y

1 − y

2

∂z

du = arcsin ydx +

x

dy + 4z3dz .

1 − y 2

7. Use the differential to estimate (0,92)3 (1,04)2 .

Solution. By a

formula

(1.6)

for

the function

f (x, y) = x3 y2 . When

x + x = 0, 92, y +

y = 1, 04 ,

x = 1 ,

y = 1 .

Hence

f (1;1) = 1,

x = 0, 92 − 1 = −0, 08,

y = 1, 04 −1 = 0, 04 . To find the partial derivatives for

function

f (x, y)

at

point

(1;1) :

∂f

= 3x 2 y 2 ,

∂f (1,1) = 3 ;

∂f

= 2x3 y ,

∂x

∂x

∂y

∂f (1,1) = 2 .

∂y

23

17

π

. Let

x +

x =

17π

,

y +

y =

π

,

x =

π

, y = 0 , when x =

π

,

z

π;

60

π

36

60

36

4

30

y =

. We have

36

π

2

π

1

∂z(π / 4,0)

z

; 0 =

sin

cos0 =

;

∂x

=sin 2x cos y

x=π / 4, =1;

4

4

2

y=0

∂z(π / 4,0) = − sin2 xsin y

x= π / 4,

= 0 ,

∂y

y=0

то

17

π

1

1

π

+ 0

π

1

π

z

π;

≈

+

=

+

≈ 0,6 .

2

30

36

2

30

60

36

а)

z = 2x4 y3 +sin 2y −

x2

;

б)

z = x2 − y 2 .

y

Solution. Since

а)

∂z

= 8x3 y3

− 2

x

,

∂z

= 6x4 y2

+ 2 cos 2y +

x2

, we have

∂x

y

∂y

y2

∂2 z

=

∂

8x

3

y

3

− 2

x

= 24x

2

y

3

−

2

and

∂x2

∂x

y

y

∂2 z

∂

4

2

x2

4

x

2

=

6x

y

+ 2 cos 2y +

= 12x

y

− 4 sin 2y − 2

,

2

2

3

∂y

∂y

y