Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
Let Fx,y=frx,y,sx,y be defined by the composition of fr,s with r=x2−y2, s=y2−x2, for any sufficiently well-behaved function fr,s. Show that y⁢Fx+x⁢Fy=0.
y Fx+x Fy
=y fr rx+fs sx+x fr ry+fs sy
=y fr 2 x+fs −2 x+x fr −2 y+fs 2 y
=fr 2 x y−fs 2 x y+fr −2 x y+fs 2 x y
=2 x y−2 x y fr+−2 x y+2 x y fs
Maple Solution - Interactive
Define rx,y and sx,y
Context Panel: Assign Name
Apply the chain rule to obtain x⁢Fx+y⁢Fy
Calculus palette: Partial-differential operator
Press the Enter key.
Context Panel: Simplify≻Simplify
y ∂∂ x fr,s+x ∂∂ y fr,s
Maple Solution - Coded
Implement the chain rule
Apply the simplify and diff commands.
y difffr,s,x +x difffr,s,y
simplifyy difffr,s,x +x difffr,s,y
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