314 CHAPTER 17. THE DERIVATIVE OF A FUNCTION OF MANY VARIABLES

Therefore, in particular,

∂ f1 ◦g∂x1

(x1,x2,x3) = (2x1x2 +2x3)x2 +1,

∂ f2 ◦g∂x3

(x1,x2,x3) = 1,∂ f2 ◦g

∂x2(x1,x2,x3) = x1 +2x2

(cos(x2

2 + x1))

.

etc.

In different notation, let

(z1

z2

)= f (u1,u2) =

(u2

1 +u2

sin(u2)+u1

). Then

∂ z1

∂x1=

∂ z1

∂u1

∂u1

∂x1+

∂ z1

∂u2

∂u2

∂x1

= 2u1x2 +1 = 2(x1x2 + x3)x2 +1.

Example 17.6.6 Let

f (u1,u2,u3) =

 z1

z2

z3

=

 u21 +u2u3

u21 +u3

2

ln(1+u2

3)

and let

g (x1,x2,x3,x4) =

 u1

u2

u3

=

 x1 + x22 + sin(x3)+ cos(x4)

x24− x1

x23 + x4

 .

Find (f ◦g)′ (x).

Df (u) =

2u1 u3 u2

2u1 3u22 0

0 0 2u3(1+u2

3)

Similarly,

Dg (x) =

 1 2x2 cos(x3) −sin(x4)

−1 0 0 2x4

0 0 2x3 1

 .

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 2x4

0 0 2x3 1



=

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

2u1−3u22 4u1x2 2u1 cosx3 −2u1 sinx4 +6u2

2x4

0 0 4 u31+u2

3x3 2 u3

1+u23

 (17.12)