Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
Example 4.3.12
If u=Fy−xx y,z−xx z, show that x2 ux+y2 uy+z2 uz=0.
Solution
Mathematical Solution
It is most convenient to define rx,y,z=y−xx y and sx,y,z=z−xx z so that x2 ux+y2 uy+z2 uz becomes x2 Fr rx+Fs sx+y2 Fr ry+Fs sy+z2 Fr rz+Fs sz. The following calculation then results from an application of the chain rule.
x2 ux+y2 uy+z2 uz
=x2 Fr rx+Fs sx+y2 Fr ry+Fs sy+z2 Fr rz+Fs sz
=x2 Fr −1x2+Fs −1x2+y2 Fr 1y2+Fs 0+z2 Fr 0+Fs 1z2
=−Fr−Fs+Fr+0+0+Fs
=−1+1Fr+−1+1Fs
=0⋅Fr+0⋅Fs
=0+0
=0
Maple Solution - Interactive
It is most convenient to define rx,y,z=y−xx y and sx,y,z=z−xx z so that x2 ux+y2 uy+z2 uz becomes x2 Fr rx+Fs sx+y2 Fr ry+Fs sy+z2 Fr rz+Fs sz.
Define r,s, and u
Context Panel: Assign Name
r=y−xx y→assign
s=z−xx z→assign
u=Fr,s→assign
Compute x2 ux+y2 uy+z2 uz and simplify the result to 0
Calculus palette: Partial-derivative operator
Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Simplify
x2 ∂∂ x u+y2 ∂∂ y u+z2 ∂∂ z u = x2⁢D1⁡F⁡y−xx⁢y,z−xx⁢z⁢−1x⁢y−y−xx2⁢y+D2⁡F⁡y−xx⁢y,z−xx⁢z⁢−1x⁢z−z−xx2⁢z+y2⁢D1⁡F⁡y−xx⁢y,z−xx⁢z⁢1x⁢y−y−xx⁢y2+z2⁢D2⁡F⁡y−xx⁢y,z−xx⁢z⁢1x⁢z−z−xx⁢z2= simplify 0
To work from first principles, obtain and simplify the following derivatives.
Separately obtain and simplify the partial derivatives ux,uy, and uz
∂∂ x u = D1⁡F⁡y−xx⁢y,z−xx⁢z⁢−1x⁢y−y−xx2⁢y+D2⁡F⁡y−xx⁢y,z−xx⁢z⁢−1x⁢z−z−xx2⁢z= simplify −D1⁡F⁡−−y+xx⁢y,−−z+xx⁢z+D2⁡F⁡−−y+xx⁢y,−−z+xx⁢zx2
∂∂ y u = D1⁡F⁡y−xx⁢y,z−xx⁢z⁢1x⁢y−y−xx⁢y2= simplify D1⁡F⁡−−y+xx⁢y,−−z+xx⁢zy2
∂∂ z u = D2⁡F⁡y−xx⁢y,z−xx⁢z⁢1x⁢z−z−xx⁢z2= simplify D2⁡F⁡−−y+xx⁢y,−−z+xx⁢zz2
Obtain and simplify rx,ry,rz,sx,sy, and sz
∂∂ x r = −1x⁢y−y−xx2⁢y= simplify −1x2
∂∂ x s = −1x⁢z−z−xx2⁢z= simplify −1x2
∂∂ y r = 1x⁢y−y−xx⁢y2= simplify 1y2
∂∂ y s = 0
∂∂ z r = 0
∂∂ z s = 1x⁢z−z−xx⁢z2= simplify 1z2
Assemble the terms of the chain rule as per the Mathematical Solution above.
Maple Solution - Coded
Initialize
Define rx,y,z.
r≔y−xx y:
Define sx,y,z.
s≔z−xx z:
Define u=Fr,s.
u≔Fr,s:
Apply the diff and simplify commands to evaluate the expression x2 ux+y2 uy+z2 uz
simplifyx2 diffu,x+y2 diffu,y+z2 diffu,z = 0
Separately obtain and simplify ux,uy, and uz
Apply the diff and simplify commands.
simplifydiffu,x = −D1⁡F⁡−−y+xx⁢y,−−z+xx⁢z+D2⁡F⁡−−y+xx⁢y,−−z+xx⁢zx2
simplifydiffu,y = D1⁡F⁡−−y+xx⁢y,−−z+xx⁢zy2
simplifydiffu,z = D2⁡F⁡−−y+xx⁢y,−−z+xx⁢zz2
Obtain and simplify the derivatives rx,ry,rz and sx,sy,sz
simplifydiffr,x = −1x2
simplifydiffs,x = −1x2
simplifydiffr,y = 1y2
simplifydiffs,y = 0
simplifydiffr,z = 0
simplifydiffs,z = 1z2
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