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Unformatted text preview: Let y = f ( x ) and s = f ( t ) and deﬁne h ( t ) = g ( f ( t )). By the defn of derivative, f ( t )f ( x ) = ( tx )[ f ( x ) + o (1)] as t → x g ( s )g ( y ) = ( sy )[ g ( y ) + o (1)] as s → y. So : h ( t )h ( x ) = g ( f ( t ))g ( f ( x )) = ( sy )[ g ( y ) + o (1)] = [ f ( t )f ( x )][ g ( f ( x )) + o (1)] = [( tx )[ f ( x ) + o (1)]][ g ( f ( x )) + o (1)] h ( t )h ( x ) tx = [ f ( x ) + o (1)][ g ( f ( x )) + o (1)] ....
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 Fall '11
 Wong
 Calculus, Derivative, Proof. Let, Proof. HW

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