474 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLES

Example 22.5.5 Let w = f (u1,u2,u3) = u21 +u3 +u2 and

g (x,y,z) =

 u1u2u3

=

 x+2yzx2 + yz2 + x

Find ∂w

∂x and ∂w∂ z

By the chain rule,

(wx,wy,wz) =(

wu1 wu2 wu3

) u1x u1y u1zu2x u2y u2zu3x u3y u3z

=

(wu1u1x +wu2u2x +wu3u3x,wu1u1y +wu2u2y +wu3u3y,

wu1u1z +wu2u2z +wu3u3z)

Note the pattern,

wx = wu1u1x +wu2u2x +wu3u3x,

wy = wu1u1y +wu2u2y +wu3u3y,

wz = wu1u1z +wu2u2z +wu3u3z.

Therefore,

wx = 2u1 (1)+1(2x)+1(1) = 2(x+2yz)+2x+1 = 4x+4yz+1

andwz = 2u1 (2y)+1(0)+1(2z) = 4(x+2yz)y+2z = 4yx+8y2z+2z.

Of course to find all the partial derivatives at once, you just use the chain rule. Thus youwould get

(wx wy wz

)=(

2u1 1 1) 1 2z 2y

2x 1 01 0 2z

=

(2u1 +2x+1 4u1z+1 4u1y+2z

)=

(4x+4yz+1 4zx+8yz2 +1 4yx+8y2z+2z

)Example 22.5.6 Let f (u1,u2) =

(u2

1 +u2sin(u2)+u1

)and

g (x1,x2,x3) =

(u1 (x1,x2,x3)u2 (x1,x2,x3)

)=

(x1x2 + x3x2

2 + x1

).

Find D(f ◦g)(x1,x2,x3).

474 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLESExample 22.5.5 Let w= f (u1,u2,u3) = ut +u3+u2 anduy x+2yzg(xyz)= | wm J=l xr+yU3 4xFind ou and ouBy the chain rule,Ujx Uly Uz(Wx, Wy, Wz) = ( Wu, Wup Wu; ) Ux, Uy UW, | =U3x U3y U3z(Wu, Ux + Way Y2x + Wu U3x5 Way Uy F Wy U2y + Wu; Uy,Wu Wz + Wu ll2z + Wu Uz)Note the pattern,Wy = Way Ux + Way Ux + Wiz 3x,Wy = Wu Uy + Way Yay + Was U3yWr = Wy Uz + Wy l2z + Wy U3z-Therefore,Wy = 2u; (1) +1 (2x) +1(1) =2(x4+2yz) + 2x41 = 404+ 4yz4+1andwe = 2uy (2y) +1 (0) +1 (2z) = 4 (w+ 2yz) y+ 2c = 4yx + By?z + 2z.Of course to find all the partial derivatives at once, you just use the chain rule. Thus youwould get1 2z 2y( wy Wy we ) = ( 2uy 1 1 ) 2x 1 01 O 2z( 2uj+2x+1 4ujz+1 4ury+2z )= (4x+4yz+1 4ex+8y22+1 4yx+8y’z+2z )2Example 22.5.6 Let f (uj ,u2) = ( sin ss) ay ) andUy \X1,%2,% XjxXo +xasim) = ( 1 (X1,%2 ) =( 1x2 ),Up (X1,%2,X3) x3 +X]Find D(f 0g) (x1,x2,%3).