Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
The composition of f⁡x,y=sin⁡2⁢x−3⁢y, with xt=t+1/t, yt=t−1/t forms the function Ft=fxt,yt. Obtain F′t by an appropriate form of the chain rule, and again by writing the rule for F explicitly. Show that the results agree.
An application of the chain rule gives
=fx x′t+fy y′t
Writing Ft=fxt,yt=−sin⁡t−5/t explicitly gives F′t=−cost−5/t⁢t2+5/t2, in agreement with the chain-rule result.
Maple Solution - Interactive
Formal statement of the relevant chain rule
Context Panel: Differentiate≻With Respect To≻t
fxt,yt→differentiate w.r.t. tD1⁡f⁡x⁡t,y⁡t⁢ⅆⅆt⁢x⁡t+D2⁡f⁡x⁡t,y⁡t⁢ⅆⅆt⁢y⁡t
It is possible to obtain notational simplifications interactively, via the Typesetting Rules Assistant in the View menu. However, this is a tedious multistep process, so will not be pursued here.
Implement the chain rule
Context Panel: Assign Function
fx,y=sin2 x−3 y→assign as functionf
Context Panel: Assign Name
Calculus palette: Partial and ordinary differential operators
Press the Enter key.
Context Panel: Evaluate at a Point≻x=X,y=Y
Context Panel: Simplify≻Simplify
∂∂ x fx,y ⅆⅆ t X+∂∂ y fx,y ⅆⅆ t Y
→evaluate at point
Obtain F′t from the explicit representation Ft=fxt,yt
Calculus palette: Differentiation operator
Context Panel: Evaluate and Display Inline
ⅆⅆ t fX,Y = −1+5t2⁢cos⁡t−5t= simplify −cos⁡t2−5t⁢t2+5t2
Maple Solution - Coded
Simplified Maple notation is available if the commands to the right are first executed.
Although the chain rule for this problem could be written as F′t=fx x′+fy y′, Maple uses the D-operator notation to express the partial derivatives fx and fy, and cannot suppress the arguments of f once suppression of arguments has been applied to x and y.
Restore the variables x and y.
Define the function f.
f≔x,y→sin2 x−3 y:
Assign xt and yt to the names X and Y, respectively.
Apply the simplify and diff commands.
simplifyD1fX,Y diffX,t+D2fX,Y diffY,t
Obtain F′t from an explicit representation of Ft
Using the diff and simplify commands, explicitly differentiate fxt,yt.
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