Chapter 4: Partial Differentiation
Section 4.11: Differentiability
Example 4.11.1
Show that the function fx,y in Table 4.11.1 has first partial derivatives everywhere.
Solution
Mathematical Solution
Let Fx,y be the rule for fx,y when x,y≠0,0.
For x,y≠0,0, the first partials of f are
fxx,y=Fxx,y=2⁢x⁢sin⁡1x2+y2−x⁢cos⁡1x2+y2x2+y2
fyx,y=Fyx,y=2⁢y⁢sin⁡1x2+y2−y⁢cos⁡1x2+y2x2+y2
At x,y=0,0, the first partials of f are
fx0,0=limh→0F0+h,0−0h=limh→0h2 sin1hh=0
fy0,0=limk→0F0,0+k−0k=limk→0k2 sin1kk=0
Consequently, the first partials are given by the following piecewise functions.
fxx,y= {2⁢x⁢sin⁡1x2+y2−x⁢cos⁡1x2+y2x2+y2x,y≠0,00x,y=0,0
and
fyx,y={2⁢y⁢sin⁡1x2+y2−y⁢cos⁡1x2+y2x2+y2x,y≠0,00x,y=0,0
Maple Solution - Interactive
Initialize
Context Panel: Assign Function
Fx,y=x2+y2⁢sin⁡1x2+y2→assign as functionF
Obtain the first partial derivatives for x,y≠0,0
Calculus palette: Partial derivative operator
Context Panel: Evaluate and Display Inline
∂∂ x Fx,y = 2⁢x⁢sin⁡1x2+y2−x⁢cos⁡1x2+y2x2+y2
∂∂ y Fx,y = 2⁢y⁢sin⁡1x2+y2−y⁢cos⁡1x2+y2x2+y2
Obtain fx0,0 and fy0,0
Calculus palette: Limit operator
limh→0F0+h,0−0h = 0
limk→0F0,0+k−0k = 0
Obtain fx0,0 and fy0,0 from first principles
Context Panel: Simplify≻Assuming Real
Context Panel: Limit For both cases, adjust the Initial Point dialog as per Figures 4.11.1(a, b), respectively. Click OK.
Figure 4.11.1(a) Limit
Figure 4.11.1(b) Limit
Fh,0h = h⁢sin⁡1h2→assuming realh⁢sin⁡1h→limit0
F0,kk = k⁢sin⁡1k2→assuming realk⁢sin⁡1k→limit0
These last two limits are zero because in each case the expressions are the product of two terms, one bounded (even through it oscillates), the other going to zero.
Maple Solution - Coded
Let F be the rule for f when x,y≠0,0.
F≔x,y→x2+y2⁢sin1x2+y2:
Apply the diff command.
diffFx,y,x = 2⁢x⁢sin⁡1x2+y2−x⁢cos⁡1x2+y2x2+y2
diffFx,y,y = 2⁢y⁢sin⁡1x2+y2−y⁢cos⁡1x2+y2x2+y2
Apply the simplify command, assuming that h and k are real.
Apply the limit command.
A≔simplifyF0+h,0−0h assuming real
h⁢sin⁡1h
limitA,h=0 = 0
B≔simplifyF0,0+k−0k assuming real
k⁢sin⁡1k
limitB,k=0 = 0
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