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Unformatted text preview: Math 53: Multivariable Calculus Worksheet 9/23/11: Differentials and the Chain Rule, 3Dified Exercise 0.1. Explain why f ( x,y ) = x + e 4 y is differentiable at (3 , 0), and give the linearization at that point. Solution. To see that f ( x,y ) is differentiable at (3 , 0), we need to find f x ( x,y ) and f y ( x,y ) and show that they are defined near (3 , 0) and continuous at (3 , 0). We have f x ( x,y ) = 1 2 x + e 4 y and f y ( x,y ) = 2 e 4 y x + e 4 y . The only places where these functions arent defined/arent continuous is when x + e 4 y = 0. At (3 , 0), we have x + e 4 y = 3 + e 4 = 4, so as long as we stay close enough to (3 , 0) we wont hit any problems. Since f x and f y are defined near (3 , 0) and continuous at (3 , 0), f ( x,y ) is differentiable at (3 , 0). The linearization of our function at ( a,b ) is L ( x,y ) = f ( a,b ) + f x ( a,b )( x a ) + f y ( a,b )( y b ) ....
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 Spring '07
 Hutchings
 Calculus, Chain Rule, Multivariable Calculus, The Chain Rule

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