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# lecture5 - Lecture 5 2.3 Chain Rule Let U and v be open...

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Unformatted text preview: Lecture 5 2.3 Chain Rule Let U and v be open sets in R n . Consider maps f : U → V and g : V R k . → Choose a ∈ U , and let b = f ( a ). The composition g ◦ f : U R k is defined by → ( g ◦ f )( x ) = g ( f ( x )). Theorem 2.9. If f is differentiable at a and g is differentiable at b , then g ◦ f is differentiable at a , and the derivative is ( Dg ◦ f )( a ) = ( Dg )( b ) ◦ Df ( a ) . (2.43) Proof. This proof follows the proof in Munkres by breaking the proof into steps. • Step 1: Let h ∈ R n − { } and h ˙=0, by which we mean that h is very close to zero. Consider Δ( h ) = f ( a + h ) − f ( a ), which is continuous, and define F ( h ) = f ( a + h ) − f ( a ) − Df ( a ) h . (2.44) a | | Then f is differentiable at a if and only if F ( h ) → 0 as h 0. → F ( h ) = Δ( h ) − Df ( a ) h , (2.45) h | | so Δ( h ) = Df ( a ) h + | h | F ( h ) . (2.46) Lemma 2.10. Δ( h ) is bounded. (2.47) h | | Proof. Define , (2.48) ∂f Df ( a ) ( a ) | = sup | ∂x i i and note that ∂f ( a ) = Df ( a ) e i , (2.49) ∂x i e i are the standard basis vectors of R n where the h = h i e i . So, we can write . If h = ( h 1 , . . . , h n ),...
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## This note was uploaded on 10/02/2010 for the course MAT unknown taught by Professor Unknown during the Fall '04 term at MIT.

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lecture5 - Lecture 5 2.3 Chain Rule Let U and v be open...

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