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M326sams_sp09

# M326sams_sp09 - Math 326A Midterm Sample 1 Consider the...

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Math 326A. Midterm Sample. 1. Consider the function f ( x, y ) = x 5 + y 5 x 3 + y 3 for ( x, y ) negationslash = (0 , 0), with the value f (0 , 0) = 0. Show that f ( x, y ) is differentiable at (0 , 0) by verifying that the partial derivatives are defined at (0 , 0) and that one of them (your choice) is continuous at (0 , 0). (Use the quotient formula to compute a partial derivative at any point ( x, y ) negationslash = (0 , 0).) Solution: ∂f ∂x | (0 , 0) = d ( f ( x, 0) dx | x =0 = d ( x 2 ) dx | x =0 = 2 x | x =0 = 0 . In the same spirit, ∂f ∂y | (0 , 0) = 0. To check for continuity of ∂f ∂y at (0 , 0), we need to see if lim ( x,y ) (0 , 0) ∂f ∂y = ∂f ∂y | (0 , 0) , i.e. if lim ( x,y ) (0 , 0) ∂f ∂y = 0. To compute the limit, we have the following formula for ∂f ∂y which is valid at points ( x, y ) negationslash = (0 , 0): ∂f ∂y = 5 y 4 ( x 3 + y 3 ) ( x 5 + y 5 )(3 y 2 ) ( x 3 + y 3 ) 2 . Because the numerator is homogeneous of degree 7 and the denominator is homogeneous of degree 6, we have lim ( x,y ) (0 , 0) ∂f ∂y = 0, which is the value of ∂f ∂y | (0 , 0) . Hence, we have established the continuity of ∂f ∂y at (0 , 0). Hence, by our theorem, f ( x, y ) is differentiable at (0 , 0).

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2. Consider the differentiable function F (
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M326sams_sp09 - Math 326A Midterm Sample 1 Consider the...

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