ChainRule

# ChainRule - THE CHAIN RULE Theorem 1 Let a,b be a non-empty...

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Unformatted text preview: THE CHAIN RULE Theorem 1. Let ( a,b ) be a non-empty 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 simple-minded approximation found in most calculus textbooks: Δ f Δ x = Δ f Δ g · Δ g Δ 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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ChainRule - THE CHAIN RULE Theorem 1 Let a,b be a non-empty...

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