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21-356 Lecture 5

# F x0 x y n f f zi x0 xi yi

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Unformatted text preview: i ) yi , yi+1 , . . . , yN ) for some "i & (0, 1) and (zi ! y( " (x ! y( . Write f (x) ! f (y) ! \$f (x0 ) · (x ! y) ' N % & !f !f = (zi ) ! (x0 ) (xi ! yi ) . ! xi ! xi i=1 Then ( N( ( |xi ! yi | |f (x) ! f (y) ! \$f (x0 ) · (x ! y)| % ( ! f !f ( ( " ( ! xi (zi ) ! ! xi (x0 )( (x ! y( . (x ! y( i=1 Since |xi #yi | \$x#y\$ " 1, we have that ( N( ( |f (x) ! f (y) ! \$f (x0 ) · (x ! y)| % ( ! f !f ( ( 0" " ( ! xi (zi ) ! ! xi (x0 )( . (x ! y( i=1 (4) Taking y = x0 and using the fact that (zi ! x0 ( " (x ! x0 ( # 0 as x # x0 , !f together with the continuity of ! xi at x0 , gives ( N( ( |f (x) ! f (x0 ) ! \$f (x0 ) · (x ! x0 )| % ( ! f !f ( ( 0" " ( ! xi (zi ) ! ! xi (x0 )( # 0, (x ! x0 ( i=1 which implies the di"erentiability of f at x0 . 11 Wednesday, January 19, 2011 The next exercise shows that the previous conditions are su!cient but not necessary for di"erentiability. Exercise 17 Let f (x, y ) := !" # 1 x2 + y 2 sin x+y 0 if (x, y ) != (0, 0) or x + y != 0, otherwise. Prove that f i...
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