Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
The composition of fx,y=3−x2−y2 with xt=t,yt=t2 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. Give a graphical interpretation of Ft.
An application of the chain rule gives
=fx x′t+fy y′t
=−2 x⋅1+−2 y⋅2 t
=−2 t⋅1+−2 t2⋅2 t
=−2 t−4 t3
Writing Ft=3−t2−t22=3−t2−t4 explicitly gives F′t=−2 t−4 t3, 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=3−x2−y2→assign as functionf
Calculus palette: Partial and ordinary differential operators
Context Panel: Evaluate at a Point≻x=t,y=t2
∂∂ x fx,y ⅆⅆ t t+∂∂ y fx,y ⅆⅆ t t2 = −4⁢t⁢y−2⁢x→evaluate at point−4⁢t3−2⁢t
Obtain F′t from an explicit representation of Ft
Write the explicit form of Ft.
Press the Enter key.
Differentiate≻With Respect To≻t
→differentiate w.r.t. t
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.
Obtain the derivative by applying the chain rule.
D1ft,t2 difft,t+D2ft,t2 difft2,t
Define the expression for Ft.
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