Chapter 4: Partial Differentiation
Section 4.2: Higher-Order Partial Derivatives
Example 4.2.6
At x,y=3,−2, obtain all third-order partial derivatives of f=x+yx−y.
Solution
Mathematical Solution
The equality of mixed partial derivatives means fxxy=fxyx=fyxx and fxyy=fyxy=fyyx. Hence, there are just four distinct third-order partial derivatives to calculate, and these appear in Table 4.2.6(a).
Derivative
at x,y
at 3,−2
fxxx
−12⁢yx−y4
24625
fxxy=fxyx=fyxx
4⁢x+2⁢yx−y4
−4625
fxyy=fyxy=fyyx
−4⁢2⁢x+yx−y4
−16625
fyyy
12⁢xx−y4
36625
Table 4.2.6(a) The distinct third-order partial derivatives of f
Maple Solution - Interactive
The partial-derivative operators in the Calculus palette can be modified to provide higher-ordered derivatives. Simply edit the notation as shown below to obtain fxxx,fyxx,fyyx, and fyyy.
Control-drag f=…
Context Panel: Assign Name
f=x+yx−y→assign
Obtain the distinct third-order partial derivatives at x,y and at 3,−2
Calculus palette: Partial-differentiation operators
Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Simplify
Context Panel: Evaluate at a Point≻x=3,y=−2
∂3∂x3 f = 6x−y3−6⁢x+yx−y4= simplify −12⁢yx−y4→evaluate at point24625
∂3∂ x2⁢∂ y f = −2x−y3+6⁢x+yx−y4= simplify 4⁢x+2⁢yx−y4→evaluate at point−4625
∂3∂ x⁢∂ y2 f = −2x−y3−6⁢x+yx−y4= simplify −4⁢2⁢x+yx−y4→evaluate at point−16625
∂3∂y3 f = 6x−y3+6⁢x+yx−y4= simplify 12⁢xx−y4→evaluate at point36625
Alternatively, to evaluate the derivatives at the point P:3,−2, define P as the list of substitutions x=3,y=−2. Had the derivatives been assigned names, they could have been referenced in the evaluation step. The choice below re-computes the derivative inside the evaluation template.
Define the substitution equations for P:3,−2
Write P=…
P=x=3,y=−2→assign
Obtain the distinct third-order partial derivatives at P:3,−2
Expression palette: Evaluation template
∂3∂x3 fx=a|f(x)P = 24625
∂3∂ x2⁢∂ y fx=a|f(x)P = −4625
∂3∂ x⁢∂ y2 fx=a|f(x)P = −16625
∂3∂y3 fx=a|f(x)P = 36625
Maple Solution - Coded
Assuming the equality of the mixed partial derivatives, define f as a function and apply the D-operator, evaluating the resulting derivatives at 3,−2.
Define the function f.
f≔x,y→x+yx−y:
Apply the D-operator to obtain the distinct third-order partial derivatives
Obtain fxxx at 3,−2.
D1,1,1f3,−2 = 24625
Obtain fxxy=fxyx=fyxx at 3,−2.
D1,1,2f3,−2 = −4625
Obtain fxyy=fyxy=fyyx at 3,−2.
D1,2,2f3,−2 = −16625
Obtain fyyy at 3,−2.
D2,2,2f3,−2 = 36625
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