# 3.6 Practice Prelim Answers - Review Problems for Prelim 3...

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Review Problems for Prelim 3 Solutions to Partial Derivatives 1. ∂f/∂x = (1 / 2)( x - y 2 ) - 1 / 2 , ∂f/∂y = (1 / 2)( x - y 2 ) - 1 / 2 ( - 2 y ) 2. ∂f/∂x = 3(3 + 2 xy 2 ) 2 (2 y 2 ), ∂f/∂y = 3(3 + 2 xy 2 ) 2 (4 xy ) 3. ∂f/∂x = 2 x x 2 + y 3 , ∂f/∂y = 3 y 2 x 2 + y 3 4. ∂f/∂x = 2 xy 3 e x 2 y , ∂f/∂y = 2 ye x 2 y + x 2 y 2 e x 2 y 5. ∂f/∂x = 2 xy ( x 2 + y 2 ) - x 2 y (2 x ) ( x 2 + y 2 ) 2 , ∂f/∂y = x 2 ( x 2 + y 2 ) - x 2 y (2 y ) ( x 2 + y 2 ) 2 6. ∂f∂x = 4 x 3 + 2 xy 2 + y 3 2 f/∂x 2 = 12 x 2 + 2 y 2 , 2 f/∂y∂x = 4 xy + 3 y 2 7. ∂f∂x = 1 y + y x 2 2 f/∂x 2 = - 2 y x 3 , 2 f/∂y∂x = - 1 y 2 + 1 x 2 8. ∂f∂x = ( 1 - y x ) e y/x 2 f/∂x 2 = y x 2 e y/x - y x 2 ( 1 - y x ) e y/x , 2 f/∂y∂x = - 1 x e y/x + 1 x ( 1 - y x ) e y/x Solutions to Definite Integrals (includes Improper Integrals) 9. 25 4 1 x dx = 2 x 1 / 2 25 4 = 10 - 4 = 6 10. Let u = 1 + x , du = dx , 0 x (1 + x ) - 5 / 2 dx = 1 (1 - u ) u - 5 / 2 du = - 2 3 u - 3 / 2 - 2 u - 1 / 2 1 = 8 / 3 11. If we let u = x 2 - 4, du = 2 x dx 3 2 8 x ( x 2 - 4) 3 dx = 5 0 4 u 3 du = u 4 5 0 = 5 4 12. 2 1 2 x 2 + 1 x dx = 2 1 2 x + 1 x dx = ( x 2 + ln x ) 2 1 = 4 + ln 2 - (1 + 0) 13. Let u = x 2 + 1, du = 2 x 2 0 2 x x 2 +1 dx = 5 1 du u = ln u | 5 1

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• Spring '07
• DURRETT

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