Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
Example 4.3.14
If w=fx,y and x=r coshs, y=r sinhs, obtain wr2−ws/r2 in terms of fx and fy.
Solution
Mathematical Solution
With ∂f∂x2 written as fx2, the following calculation applies the appropriate form(s) of the applicable chain rule.
wr2−ws/r2
=fx xr+fy yr2−fx xs+fy ys/r2
=fx coshs+fy sinhs2−fx r sinhs+fy r coshs/r2
=fx2 cosh2s+fy2 sinh2s +2 sinh2scosh2sfxfy −fx2 sinh2s+fy2 cosh2s +2 sinh2scosh2sfxfy
=cosh2s−sinh2sfx2−cosh2s−sinh2sfy2
=1⋅fx2−1⋅fy2
=fx2−fy2
Maple Solution - Interactive
Define r,s, and u
Context Panel: Assign Name
x=r coshs→assign
y=r sinhs→assign
w=fx,y→assign
Compute wr2−ws/r2 and simplify the result
Calculus palette: Partial-derivative operator
Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Simplify
∂∂ r w2−∂∂ s w/r2 = D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢cosh⁡s+D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢sinh⁡s2−D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢r⁢sinh⁡s+D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢r⁢cosh⁡s2r2= simplify D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s2−D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s2
To work from first principles, obtain and simplify the following derivatives.
Separately obtain and simplify the partial derivatives ux,uy, and uz
∂∂ r w = D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢cosh⁡s+D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢sinh⁡s
∂∂ s w = D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢r⁢sinh⁡s+D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢r⁢cosh⁡s
Obtain xr,xs and yr,ys
∂∂ r x = cosh⁡s
∂∂ r y = sinh⁡s
∂∂ s x = r⁢sinh⁡s
∂∂ s y = r⁢cosh⁡s
Assemble the terms of the chain rule as per the Mathematical Solution above.
Maple Solution - Coded
Initialize
Define xr,s.
x≔r coshs:
Define yr,s.
y≔r sinhs:
Define w=fx,y.
w≔fx,y:
Apply the diff and simplify commands to evaluate the expression wr2−ws/r2
simplifydiffw,r2− diffw,s/r2 = D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s2−D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s2
Obtain wr and ws
Apply the diff command.
diffw,r = D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢cosh⁡s+D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢sinh⁡s
diffw,s = D1⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢r⁢sinh⁡s+D2⁡f⁡r⁢cosh⁡s,r⁢sinh⁡s⁢r⁢cosh⁡s
Obtain the derivatives xr,xs and yr,ys
diffx,r = cosh⁡s
diffy,r = sinh⁡s
diffx,s = r⁢sinh⁡s
diffy,s = r⁢cosh⁡s
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