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Unformatted text preview: deﬁned. On the
other hand, if were diﬀerentiable at the point (0 0 ), both partial derivatives would
exist. Thus is not diﬀerentiable at = 0, = 0 = 0.
3 (c) (d) = 0 = 0 and = 0. Since there is a symmetry (
similar to the previous one and is not diﬀerentiable at ( )= (
0 0). ), this case is =
= 0. Here ( 0) ≡ 0, so (0 0) = 0 and, similarly,
diﬀerentiability we need to see whether
( − (0 0) − lim )→(0 0) ( 2 (0 0) ) 2 + 
= lim
( )→(0 0) (0 0) = 0. For 
2 + 2 =0 Using what we learned in the lectures, i.e. that
 ≤ (  ≤ ( ) ) we see that 0≤  
2 + 2 = )·(
(
) (
 · 
≤
(
) )
=( ) →0 as ( ) → (0 0) (the ﬁnal part, that (
) → 0 as (
) → (0 0), you can easily prove with the ε − δ
deﬁnition). Hence, by the squeeze theorem

lim )→(0 0) ( and we see that ( 3 
2 + 2 =0 ) is indeed diﬀerentiable at (0 0). Extra problems on derivatives and diﬀerentiability E
Let (
) =  . Recall from last tutorial that
that, however, is continuous but not diﬀerentiable at (0 0).
S We ﬁrst show that ( (0 0) = (0 0) = 0. Prove ) is continuous at (0 0), i.e. we show that
lim
( )→(0 0) ( ) = (0 0) in other words, we show that
 lim
( )→(0 0) 4 =0 Using what we learned in the lectures, i.e. that
 ≤ (  ≤ ( ) ) we see that
0≤  =  · ≤ )·( ( )=( ) Since trivially (you can prove this easily with the εδ deﬁnition), (
(0 0), we get by the squeeze theorem that
 lim )→ =0 )→(0 0) ( ) → 0 as ( i.e. (
) is continuous at (0 0).
Now to show that is not diﬀerentiable at (0 0), we note that by deﬁnition, diﬀerentiability means that (since (0 0) = (0 0) = 0 as we calculated last tutorial)

lim
( )→(0 0) −0 −0 
2 + 
= 2 lim
( )...
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This document was uploaded on 02/10/2014.
 Spring '13

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