10.12 - Section 11.5 The chain rule(concluded Implicit...

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Section 11.5: The chain rule (concluded) Implicit differentiation revisited: Suppose ( a , b ) is a point on the curve C = {( x , y ): F ( x , y ) = 0} (e.g., when the function F ( x , y ) is x 2 + y 2 – 1, the curve C is the unit circle), so that F ( a , b ) = 0. We’d like to know that there exists a piece of the curve that is the graph of a function and contains the point ( a , b ). That is, we’d like to know that there exists a function f defined on some interval ( a r , a + r ) (with r > 0) such that f ( a ) = b and F ( x , f ( x )) = 0 for all x in ( a r , a + r ). But how can we be sure that the equation F ( x , y ) = 0 determines y as a function of x in the vicinity of a particular point ( a , b ) satisfying F ( a , b ) = 0? In fact, we’ve seen that this doesn’t work for the points ( ± 1,0) on the circle x 2 + y 2 =1! The Implicit Function Theorem tells us that the equation F ( x , y ) = 0 determines y as a function of x in the vicinity of a particular point (

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10.12 - Section 11.5 The chain rule(concluded Implicit...

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