Math 425/575 SELECTED PRACTICE PROBLEMS WITH SOLUTIONS Winter
2007
11. Let f(x, y) = x3/(x2+y2) for (x, y)6= (0,0) and f(0,0) = 0. Prove that fis not
differentiable at (0,0).
Solution. Suppose that fis differentiable at 0. (Note that the function is continuous, so we
cannot get a contradiction this way.) It it is easy to see that D1f(0) = 1 and D2f(0) = 0; thus
the total derivative would have to be f0(0)(x, y) = (1,0)·(x, y ) = x. Now there are several ways
to proceed and get a contradiction. For example, compute the directional derivative f0(0,u)
for u= (1, a):
f0(0,u) = lim
h→0
f(h, ah)
h= lim
h→0
h3
(h2+a2h2)h=1
1 + a2.
On the other hand, f0(0,u) = ∇f(0)·u= (1,0) ·(1, a) = 1 which is a contradiction when
a6= 1.
12. Consider a function fon R2such that f(0,0) = 0 and
f(x, y) = x2y6
(x2+y4)2,for (x, y)6= (0,0).
(a) Prove that fis differentiable at every point.
(b) Prove that fis not continuously differentiable.
(a) Clearly, fis continuously differentiable everywhere, except possibly at the origin. It is
easy to see that the partial derivatives at 0= (0,0) are equal to zero, hence f0(0) = 0, if it
exists. By the definition of total derivative, all we need to check is that
f(x, y)/k(x, y)k → 0 as (x, y )→0.
We have f(x, y)
k(x, y)k=x2y6
(x4+ 2x2y4+y8)√x2+y2)≤(1/2) y2
√y2= (1/2)|y| → 0,
as (x, y)→0, q.e.d.
Another solution (sketch): prove that D2fis continuous at 0and apply Theorem 12.11.
(b)
D1f(x, y) = 2xy6
(x2+y4)2−2x2y6·2x
(x2+y4)3, D1f(ay2, y) = 2a
(a2+ 1)2−4a3
(a2+ 1)3=2a(1 −a2)
(a2+ 1)3,
which is a constant, not equal to zero for a6= 1. Thus, D1fis discontinuous at 0(in view of
D1f(0) = 0).
13. Let abe a fixed vector in Rn. Consider the function F:Rn→Rngiven by
F(x) = (a·x)x.
Compute the total derivative F0(x) (it should be expressed as a linear function).
