Tutorial04-solutions

# Thus is not dierentiable at 0 0 0 3 c d 0

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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 ε-δ 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.

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