476 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLES

Then by the chain rule, D(f ◦g)(x)=Df (u)Dg (x) where u= g (x) as described above.Thus D(f ◦g)(x) = 2u1 u3 u2

2u1 3u22 0

0 0 2u3(1+u2

3)

 1 2x2 cos(x3) −sin(x4)

−1 0 0 2x40 0 2x3 1



=

 2u1 −u3 4u1x2 2u1 cosx3 +2u2x3 −2u1 sinx4 +2u3x4 +u22u1 −3u2

2 4u1x2 2u1 cosx3 −2u1 sinx4 +6u22x4

0 0 4 u31+u2

3x3 2 u3

1+u23

 (22.13)

where each ui is given by the above formulas. Thus ∂ z1∂x1

equals

2u1 −u3 = 2(x1 + x2

2 + sin(x3)+ cos(x4))−(x2

3 + x4)

= 2x1 +2x22 +2sinx3 +2cosx4 − x2

3 − x4.

while ∂ z2∂x4

equals

−2u1 sinx4 +6u22x4 =−2

(x1 + x2

2 + sin(x3)+ cos(x4))

sin(x4)+6(x2

4 − x1)2

x4.

If you wanted ∂z∂x2

it would be the second column of the above matrix in 22.13. Thus ∂z∂x2

equals ∂ z1∂x2∂ z2∂x2∂ z3∂x2

=

 4u1x24u1x2

0

=

 4(x1 + x2

2 + sin(x3)+ cos(x4))

x24(x1 + x2

2 + sin(x3)+ cos(x4))

x20

I hope that by now it is clear that all the information you could desire about various partialderivatives is available and it all reduces to matrix multiplication and the consideration ofentries of the matrix obtained by multiplying the two derivatives.

22.6 Exercises1. Let z = f (x1, · · · ,xn) be as given and let xi = gi (t1, · · · , tm) as given. Find ∂ z

∂ tiwhich

is indicated.

(a) z = x31 + x2, x1 = sin(t1)+ cos(t2) ,x2 = t1t2

2 . Find ∂ z∂ t1

(b) z = x1x22, x1 = t1t2

2 t3,x2 = t1t22 . Find ∂ z

∂ t1.

(c) z = x1x22, x1 = t1t2

2 t3,x2 = t1t22 . Find ∂ z

∂ t1.

(d) z = x1x22, x1 = t1t2

2 t3,x2 = t1t22 . Find ∂ z

∂ t3.

(e) z = x21x2

2, x1 = t1t22 t3,x2 = t1t2

2 . Find ∂ z∂ t2

.

(f) z = x21x2 + x2

3, x1 = t1t2,x2 = t1t2t4,x3 = sin(t3). Find ∂ z∂ t2

.

(g) z = x21x2 + x2

3, x1 = t1t2,x2 = t1t2t4,x3 = sin(t3). Find ∂ z∂ t3

.