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2s4dLg-04

2s4dLg-04 - MAT 473 9 4 Chain Rule Partial Derivatives...

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MAT 473 9 4. Chain Rule. Partial Derivatives. Proposition 4.1 (Chain Rule) . Suppose f : U R n R m is di ff erentiable at x U and g : V R m R is di ff erentiable at f ( x ) V . Then g f is di ff erentiable at x , with ( g f ) ( x ) = g ( f ( x )) f ( x ) . Proof. Put y = f ( x ) and choose s, W s and t, W t as in Lemma 3.6 such that f ( x + h ) = f ( x ) + f ( x )( h ) + s ( h ) h for all h W s , g ( y + k ) = g ( y ) + g ( y )( k ) + t ( k ) k for all k W t , lim h 0 s ( h ) = 0 , and lim k 0 t ( k ) = 0 . Now let k ( h ) = f ( x )( h ) + s ( h ) h ; certainly lim h 0 k ( h ) = 0 , so we can choose > 0 such that k ( h ) W t whenever h < . Thus for all h W = W s B ( 0 ), we have ( g f )( x + h ) = g ( f ( x ) + f ( x )( h ) + s ( h ) h ) = g ( y + k ( h )) = g ( y ) + g ( y )( k ( h )) + t ( k ( h )) k ( h ) = g ( f ( x )) + g ( f ( x ))( f ( x )( h ) + s ( h ) h ) + t ( k ( h )) k ( h ) = ( g f )( x ) + g ( f ( x ))( f ( x )( h )) + g ( f ( x ))( s ( h )) h + t ( k ( h )) k ( h ) h h . Now clearly lim h 0 g ( f ( x ))( s ( h )) = 0 . Moreover, 0 t ( k ( h )) k ( h ) h = t ( k ( h )) f ( x )( h ) + s ( h ) h h t ( k ( h )) f ( x ) h + s ( h ) h h = t ( k ( h )) ( f ( x ) + s ( h ) ) 0 as h 0 , so lim h 0 t ( k ( h )) k ( h ) h = 0 . Since h g ( f ( x ))( f ( x )( h )) = ( g ( f ( x )) f ( x )) ( h ) is linear, another application of Lemma 3.6 completes the proof. Definition 4.2. For f : U R n R m , and for 1 i m , define f i : U R by f i ( x ) = e i , f ( x ) = f ( x ) i for all x R n . We call f i the i th

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2s4dLg-04 - MAT 473 9 4 Chain Rule Partial Derivatives...

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