“JUST THE MATHS” UNIT NUMBER 14.2 PARTIAL DIFFERENTIATION 2 (Partial derivatives of order higher than one) by A.J.Hobson 14.2.1 Standard notations and their meanings 14.2.2 Exercises 14.2.3 Answers to exercises UNIT 14.2 - PARTIAL DIFFERENTIATION 2 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) ∂2z ∂ , which means 2 ∂x ∂x ∂z ; ∂x ∂2z ∂ , which means 2 ∂y ∂y ∂z ; ∂y ! (ii) ! (iii) ∂2z ∂ , which means ∂x∂y ∂x ∂z ; ∂y ∂2z ∂ , which means ∂y∂x ∂y ∂z . ∂x ! (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 ∂3z ∂ ∂ , which means ∂x∂y 2 ∂x ∂y " 1 ∂z ∂y !# and ∂4z ∂ , which means 2 2 ∂x ∂y ∂x " ∂ ∂ ∂x ∂y ∂z ∂y !#! . EXAMPLES Determine all the first and second order partial derivatives of the following functions: 1. z = 7x3 − 5x2 y + 6y 3 . Solution ∂z ∂x = 21x2 − 10xy; ∂z ∂y ∂2z ∂x2 = 42x − 10y; ∂2z ∂y 2 ∂2z ∂y∂x = −5x2 + 18y 2 ; = 36y; ∂2z ∂x∂y = −10x; = −10x. 2. z = y sin x + x cos y. Solution ∂z ∂x ∂2z ∂x2 ∂z ∂y = y cos x + cos y; ∂2z ∂y 2 = −y sin x; ∂2z ∂y∂x = sin x − x sin y; = −x cos y; ∂2z ∂x∂y = cos x − sin y; 2 = cos x − sin y. 3. z = exy (2x − y). Solution ∂z ∂x ∂2z ∂x2 ∂z ∂y = exy [y(2x − y) + 2] = exy [2xy − y 2 + 2]; ∂2z ∂y 2 = exy [y(2xy − y 2 + 2) + 2y] = exy [x(2x2 − xy − 1) − x] = exy [2xy 2 − y 3 + 4y]; ∂2z ∂y∂x = exy [x(2x − y) − 1] = exy [2x2 − xy − 1]; = exy [2x3 − x2 y − 2x]; ∂2z ∂x∂y = exy [x(2xy − y 2 + 2) + 2x − 2y] = exy [y(2x2 − xy − 1) + 4x − y] = exy [2x2 y − xy 2 + 4x − 2y]; = exy [2x2 y − xy 2 + 4x − 2y]. 14.2.2 EXERCISES 1. Determine all the first and second order partial derivatives of the following functions: (a) z = 5x2 y 3 − 7x3 y 5 ; (b) z = x4 sin 3y. 2. Determine all the first and second order partial derivatives of the function w ≡ z 2 exy + x cos(y 2 z). 3. If ! x , z = (x + y) ln y show that x2 2 ∂2z 2∂ z = y . ∂x2 ∂y 2 3 4. If z = f (x + ay) + F (x − ay), show that ∂2z 1 ∂2z = 2 2. ∂x2 a ∂y 14.2.3 ANSWERS TO EXERCISES 1. (a) The required partial derivatives are as follows: ∂z ∂x = 10xy 3 − 21x2 y 5 ; ∂z ∂y ∂2z ∂x2 = 10y 3 − 42xy 5 ; ∂2z ∂y 2 ∂2z ∂y∂x = 15x2 y 2 − 35x3 y 4 ; = 30x2 y − 140x3 y 3 ; ∂2z ∂x∂y = 30xy 2 − 105x2 y 4 ; = 30xy 2 − 105x2 y 4 . (b) The required partial derivatives are as follows: ∂z ∂x ∂2z ∂x2 ∂z ∂y = 4x3 sin 3y; ∂2z ∂y 2 = 12x2 sin 3y; ∂2z ∂y∂x = 3x4 cos 3y; = −9x4 sin 3y; ∂2z ∂x∂y = 12x3 cos 3y; = 12x3 cos 3y. 2. The required partial derivatives are as follows: ∂w = yz 2 exy + cos(y 2 z); ∂x ∂w = z 2 xexy − 2xyz sin(y 2 z); ∂y ∂w = 2zexy − xy 2 sin(y 2 z); ∂z ∂2w ∂2w ∂2w 2 2 xy 2 2 xy 2 2 2 2 = y z e ; = z x e −2xz sin(y z)+4xy z cos(y z); = 2exy −xy 4 cos(y 2 z); ∂x2 ∂y 2 ∂z 2 4 ∂2w ∂2w = = z 2 exy + z 2 xyexy − 2yz sin(y 2 z); ∂x∂y ∂y∂x ∂2w ∂2w = = 2zxexy − 2xy sin(y 2 z) − 2xy 3 z cos(y 2 z); ∂y∂z ∂z∂y ∂2w ∂2w = = 2zyexy − y 2 sin(y 2 z). ∂z∂x ∂x∂z 3. ∂2z 1 y ∂2z 1 x = − and = − + 2. 2 2 2 ∂x x x ∂y y y 4. ∂2z ∂2z 00 00 = f (x + ay) + F (x − ay) and = a2 f 00 (x + ay) + a2 F 00 (x − ay). ∂x2 ∂y 2 5