Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
If u is a function of r=x2+y2+z2, show that ux2+uy2+uz2=dudr2.
It is most convenient to define rx,y,z=x2+y2+z2 and u=gr. The following calculation then results from an application of the chain rule.
=g′ xr2+g′ yr2+g′ zr2
Maple Solution - Interactive
Define r and u
Context Panel: Assign Name
Compute ux2+uy2+ uz2
Calculus palette: Partial-derivative operator
Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Simplify
∂∂ x u2+ ∂∂ y u2+∂∂ z u2 = D⁡g⁡x2+y2+z22⁢x2x2+y2+z2+D⁡g⁡x2+y2+z22⁢y2x2+y2+z2+D⁡g⁡x2+y2+z22⁢z2x2+y2+z2= simplify D⁡g⁡x2+y2+z22
To work from first principles, obtain and simplify the following derivatives.
Obtain and square the partial derivatives ux,uy, and uz
∂∂ x u = D⁡g⁡x2+y2+z2⁢xx2+y2+z2
∂∂ x u2 = D⁡g⁡x2+y2+z22⁢x2x2+y2+z2
∂∂ y u = D⁡g⁡x2+y2+z2⁢yx2+y2+z2
∂∂ y u2 = D⁡g⁡x2+y2+z22⁢y2x2+y2+z2
∂∂ z u = D⁡g⁡x2+y2+z2⁢zx2+y2+z2
∂∂ z u2 = D⁡g⁡x2+y2+z22⁢z2x2+y2+z2
The sum of the squares of the partial derivatives is then
where the name g reverts back to u.
Maple Solution - Coded
Apply the diff and simplify commands to evaluate the expression ux2+uy2+ uz2
simplifydiffu,x2+ diffu,y2+diffu,z2 = D⁡g⁡x2+y2+z22
(Note the use of ux2 for ux2, etc.)
Separately obtain ux,uy, and uz and their squares
Apply the diff and
diffu,x = D⁡g⁡x2+y2+z2⁢xx2+y2+z2
diffu,x2 = D⁡g⁡x2+y2+z22⁢x2x2+y2+z2
diffu,y = D⁡g⁡x2+y2+z2⁢yx2+y2+z2
diffu,y2 = D⁡g⁡x2+y2+z22⁢y2x2+y2+z2
diffu,z = D⁡g⁡x2+y2+z2⁢zx2+y2+z2
diffu,z2 = D⁡g⁡x2+y2+z22⁢z2x2+y2+z2
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