Solutions

UNIT 14.2

- PARTIAL DIFFERENTIATION2

PARTIAL DERIVATIVES OF THE SECOND AND HIGHER ORDERS 14.2.1 STANDARD

NOTATIONS

AND THEIR

MEANINGS

In Unit 14.1, the partial derivatives encountered are known as partial derivatives of the first order; that is, the dependent variable was differentiated only once with respect to each independent variable. But a partial derivative will, in general contain all of the independent variables, suggesting that we may need to differentiate again with respect to any of those variables. For example, in the case where a variable, z, is a function of two independent variables, x and y, the possible partial derivatives of the second order are (i) (j2z. 8 8X2' whIch means 8x

8Z

() 8x

;

(ii)

(Hi)

(iv)

The last two can be shown to give the same result for all elementary functions likely to be encountered in science and engineering. Note: Occasionally, it may be necessary to use partial derivatives of order higher than two, as illustrated, for example, by

and 84Z

'

8x28y2

EXAMPLES Determine all the first and second order partial derivatives of the following functions: 1.

Solution 8z 8x ~ 2lx2 - lOx y 'C,,'J~

82; 8x

= 42x

82Z 8y8x

8z 8y

= -5X2

+ l8 y2.,

82Z

- lOy;

8iJi = 36y; 82Z

= -1 Ox;

= -lOx.

8x8y

2. z

= ysinx + x cosy.

Solution 8z

8x

= y cosx+cos

82Z'

8X2 = 82Z

8y8x

'

y'

.

-ysmx;

= cos x - sin y;

8z 8y

= sin x

- xsin y ',

82Z

8iJi= -xcosy; 82Z

8x8y = cosX - sin y.

and {)4Z {)X2{)y2' which means

~ (:X[~ (~;)]).

EXAMPLES Determine all the first and second order partial derivatives of the following functions: 1.

Solution 8z ~ 21x2 -lOx y 'c.,'J~

8x

82Z

8X2= 42x

8z = -5X2 + 18y 2., 8y

- lOyj 82Z

8x8y

= -lOx.

2.

z = ysinx + xcosy. Solution ' 8x = Y cosx + cosy ,

8z

82Z" 8X2

82Z

8y8x

=

.

= sin X - xsin y ,

82Z

-ysmXj

= cos X -

8z 8y 7fiJ

sin yj

'

= -x cosy;

3.

Solution ~~ = eXY[y(2x- y) + 2] = eXY[2xy- y2 + 2]j

~~

82Z

8y2

=

eXY[x(2x

= eXY[2x2

= eXY [x (2X2 -

= eXY[x(2xy - y2 + 2) + 2x - 2y] = eXY[2x2y -

:;;Y

= eXY[y(2X2

y) - 1]

xy - 1) - x]

= eXY[2x3 -

:;;X

-

- xy - 1];

X2y

-

2x];

- xy - 1) + 4x - y]

= eXY[2x2y- Xy2+ 4x - 2y].

Xy2 + 4x - 2Y]j

14.2.2 EXERCISES 1. Determine all the first and second order partial derivatives of the following functions: (a)

(b) z 2. Determine

= X4 sin3y.

all the first and second order partial derivatives of the function

~. 3. If

show that

z = (x + y) Ih (~), ~

.~

,

4. If

-

z = f(x + ay) + F(x

ay),

show that E]2z OX2

1 (]2z

-

a2 oy2

.

14.2.3 ANSWERS TO EXERCISES 1. (a) The required partial derivatives are as follows:

lPz = 30x y 2 8y8x

lPz = 30x y 2 - 105x2y4. - 105x2y 4.' 8x8y

(b) The required partial derivatives are as follows: 8z 8x

= 4x3 sin 3yj

8z 8y

= 3X4 cos 3yj

82Z 4 . 8il = -9x sm 3yj

lPz 8x8y

2. The required partial derivatives are as follows:

= 12x3cos3y.

fPw

82w

8z8x = 8x8z = 2zyexy - y2 sin(y2z). 3.

4. 82Z

82Z

8X2 = f"(x + ay) + F"(X - ay) and 8y2 = a2f"(x + ay) + a2F"(X - ay).