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practice1-solution - Math 3210 practice problems for first...

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Unformatted text preview: Math 3210 practice problems for first prelim Fall 2009 Some hints and solutions This is not a detailed solution sheet, but rather a collection of hints and summary answers intended to help you along in preparation for the exam. 1. dg = 2 f ( x ) cos ( f ( x ) 2 ) d f . 2. As in several of the other problems, the trick here is to restate the problem in such a way that you can apply the (multivariable) chain rule. Define g : R 2 → R 2 by g x y ¶ = y x ¶ . Then f ( x , y ) =- f ( y , x ) is equivalent to ( f ◦ g )( x , y ) =- f ( x , y ) . This implies D ( f ◦ g )( x , y ) =- D f ( x , y ) , and therefore, by the chain rule, D f ( g ( x , y )) Dg ( x , y ) =- D f ( x , y ) . Therefore, using Dg ( x , y ) = 0 1 1 0 ¶ , D f ( x , y ) = ∂ f ∂ x ( x , y ) , ∂ f ∂ y ( x , y ) ¶ , we get ∂ f ∂ x ( g ( x , y )) , ∂ f ∂ y ( g ( x , y )) ¶ 0 1 1 0 ¶ =- ∂ f ∂ x ( x , y ) , ∂ f ∂ y ( x , y ) ¶ , which leads to the correct answer....
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