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Unformatted text preview: 236 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION
but if we try to calculate d(0,0) f (− ) for the vector − = (1, m) using
(f (t, mt) − f (0, 0))
t→0 t d(0,0) f (1, m) = lim then, since along the line y = mx the function has a constant value
f (t, mt) = m
1 + m2 —which is nonzero if m is—but f (0, 0) = 0, we see that the limit above
does not exist:
(f (t, mt) − f (0, 0)) = lim
1 + m2 diverges, and the diﬀerential cannot be evaluated along the vector
− = − + m− if m = 0. In fact, we saw before that this function is not
ı continuous at the origin, which already contradicts diﬀerentiability, by
Another example, this time one which is continuous at the origin, is √ 2xy
if (x, y ) = (0, 0),
x 2 +y 2
f (x, y ) =
at (0, 0) This function is better understood when expressed in polar coordinates,
where it takes the form
2r 2 cos θ sin θ
= 2r cos θ sin θ f (r cos θ, r sin θ ) = = r sin 2θ.
From this we see that along the line making angle θ with the x-axis,
f (x, y ) is a constant (sin 2θ ) times the distance from the origin:
geometrically, the graph of f (x, y ) over this line is itself a line through the
origin of slope m = sin 2θ . Along the two coordinate axes, this slope is
zero, but for example along the line y = x (θ = π/4), the slope is
sin π/2 = 1. So this time, the function deﬁned by Equation (3.6) (without
asking about diﬀerentiability) exists at the origin, but it is not linear
(since again it is zero on each of the standard basis elements − and − ).
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
- Fall '08