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Unformatted text preview: THE CHAIN RULE Theorem 1. Let ( a,b ) be a nonempty open interval, and g : ( a,b ) R a function. Suppose x ( a,b ) , and let ( c,d ) be an open interval containing g ( x ) . Let f : ( c,d ) R be another function. If g is differentiable at x , and f is differentiable at g ( x ) , then f g is differentiable at x , and ( f g ) ( x ) = f ( g ( x )) g ( x ) . In order to prove this theorem, we follow our nose and refer to the simpleminded approximation found in most calculus textbooks: f x = f g g x . Take h small (in absolute value), and consider 1 h [ f ( g ( x + h )) f ( g ( x ))] . We want the numerator to look like f ( g ( x ) + k ) f ( g ( x )) for a small k , so we set g ( x ) + k = g ( x + h ) , and define k ( h ) = g ( x + h ) g ( x ) . The idea is then to factor f ( g ( x + h )) f ( g ( x )) h = f ( g ( x ) + k ( h )) f ( g ( x )) k ( h ) k ( h ) h . Well, k ( h ) /h = 1 h [ g ( x + h ) g ( x )] which tends to g ( x ) as h . Also, since g is differ entiable at x , it is continuous there, and so lim h k ( h ) = lim h [ g ( x + h ) g ( x )] = 0 . Hence, if h n is any sequence tending to , the sequence k n = k ( h n ) tends to , and so since f is differentiable at g ( x ) we have f ( g ( x ) + k n ) f ( g ( x )) k n f ( g ( x )) as n . The limit theorem for products thus completes the proof of the Chain Rule.....
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 Fall '10
 Prof.KatrinWehrheim
 Chain Rule, The Chain Rule

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