22.5. THE CHAIN RULE 473

From 22.12, ∂ z∂ t =

∂ z∂u

∂u∂ t +

∂ z∂v

∂v∂ t = vcos(uv)sin(x)+uscos(uv) . Here y1 = u,y2 = v, t =

xk. Also,

∂ z∂x

=∂ z∂u

∂u∂x

+∂ z∂v

∂v∂x

= vcos(uv) t cos(x)+ussec2 (x)cos(uv) .

Clearly you can continue in this way, taking partial derivatives with respect to any of theother variables.

Example 22.5.4 Let w = f (u1,u2) = u2 sin(u1) and u1 = x2y + z,u2 = sin(xy). Find∂w∂x ,

∂w∂y , and ∂w

∂ z .

The derivative of f is of the form (wx,wy,wz) and so it suffices to find the derivative off using the chain rule. You need to find D f (u1,u2)Dg (x,y,z) where

g (x,y) =(

x2y+ zsin(xy)

).

Then

Dg (x,y,z) =(

2xy x2 1ycos(xy) xcos(xy) 0

).

Also D f (u1,u2) = (u2 cos(u1) ,sin(u1)). Therefore, the derivative is

D f (u1,u2)Dg (x,y,z) = (u2 cos(u1) ,sin(u1))

(2xy x2 1

ycos(xy) xcos(xy) 0

)

=(2u2 (cosu1)xy+(sinu1)ycosxy,u2 (cosu1)x2 +(sinu1)xcosxy,u2 cosu1)

= (wx,wy,wz)

Thus

∂w∂x

= 2u2 (cosu1)xy+(sinu1)ycosxy = 2(sin(xy))(cos(x2y+ z

))xy

+(sin(x2y+ z

))ycosxy.

Similarly, you can find the other partial derivatives of w in terms of substituting in for u1and u2 in the above. Note

∂w∂x

=∂w∂u1

∂u1

∂x+

∂w∂u2

∂u2

∂x.

In fact, in general if you havew = f (u1,u2)

and g (x,y,z) =(

u1 (x,y,z)u2 (x,y,z)

), then D( f ◦g)(x,y,z) is of the form

(wu1 wu2

)( u1x u1y u1zu2x u2y u2z

)=

(wu1ux +wu2 u2x wu1uy +wu2 u2y wu1uz +wu2u2z

).