exam2_2009_sol

# 6 find the following derivatives do not simplify page

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6. Find the following derivatives. Do not simplify. Page 3

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Math 171: Exam 2 (a) [7 points] Find d dx p x + (5 x 2 + 1) 100 . Solution: Using the chain rule (twice): d dx ( x + (5 x 2 + 1) 100 ) 1 / 2 = 1 2 ( x + (5 x 2 + 1) 100 ) - 1 / 2 1 + 100(5 x 2 + 1) 99 (10 x ) . (b) [7 points] d dx (2 x + 1) p x 3 x 2 + 1 . Solution: Using the quotient rule and the product rule: d dx (2 x + 1) p x 3 x 2 + 1 = (3 x 2 + 1) 2 p x + (2 x + 1)( 1 2 x - 1 / 2 ) - (2 x + 1) p x (6 x ) (3 x 2 + 1) 2 . 7. (a) [6 points] Give the definition of “ f ( x ) is continuous at x = c .” Solution: f ( x ) is continuous at x = c if lim x ! c f ( x ) = f ( c ). (More precisely: (1) c is in the domain of f ; (2) lim x ! c f ( x ) exists; and (3) lim x ! c f ( x ) = f ( c ).) (b) [6 points] Explain why f ( x ) = x 3 - x 2 - 5 p x + 1 has a root in the interval [0 , 3]. Solution: Note that f (0) = 0 - 0 - 5 p 1 = - 5 < 0 and f (3) = 3 3 - 3 2 - 5 p 4 = 27 - 9 - 10 = 8 > 0. Furthermore, f is continuous on [0 , 3]. So be the Intermediate Value Theorem , there is a c in the interval (0 , 3) such that f ( c ) = 0. 8. (a) [6 points] Find the tangent line to the graph of x 2 y 2 - 2 = y at (1 , 2). Solution: Di erentiating both sides in terms of x we get 2 xy 2 + 2 x 2 y dy dx = dy dx . Rearranging, 2 xy 2 = dy dx (1 - 2 x 2 y ). Therefore, dy dx = 2 xy 2 1 - 2 x 2 y . At (1 , 2), dy dx = - 8 3 . The tangent line has equation y - 2 = - 8 3 ( x - 1). (b) [6 points] If dy dx = x y 2 , find d 2 y dx 2 . Be sure to express your answer only in terms of x and y .
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• Fall '07
• GOMEZ,JONES

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