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Unformatted text preview: Section 11.3: Partial derivatives (concluded) Clairaut’s Theorem says that f 12 ( a , b ) and f 21 ( a , b ) must be equal , as long as the functions f 12 and f 21 are not only defined at ( a , b ) but are also defined in a neighborhood of the point ( a , b ) and in fact are continuous in a (diskshaped) neighborhood of ( a , b ). Suppose that f ( x , y ) is continuous everywhere. Assume that f x (1,1) = 2, f y (1,1) = –2, and f xy (1,1) = 3. Is it possible to compute f yx (1,1) from this information alone? … ..?.. No; in order to be able to apply Clairaut’s Theorem, we would need to know that f xy and f yx are continuous in a disk containing (1,1). [Go through proof of Clairaut’s Theorem on page A23.] In the case where f 12 and f 21 are both continuous (and hence by Clairaut’s Theorem equal to one another), the proof gives us an alternative way to think about the common value of f 12 ( a , b ) and f 21 ( a , b ): it’s the limit of (*) [ f ( a + h , b + h ) – f ( a + h , b ) – f ( a , b + h ) + f ( a , b )] / h 2 as h goes to 0. [Draw picture.] Example: Let f ( x , y ) = xy . Then f 1 ( x , y ) = … ..?.....
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 Fall '11
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 Derivative

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