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
<< Previous Example Section 4.3
Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2021. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
What kind of issue would you like to report? (Optional)