2 h h(2 x h 1(2 x 1 = lim h 2(2 x h 1(2 x 1 = 2(2 x 0

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Unformatted text preview: - 2 h h (2( x + h )- 1)(2 x- 1) = lim h !- 2 (2( x + h )- 1)(2 x- 1) =- 2 (2( x + 0)- 1)(2 x- 1) =- 2 (2 x- 1) 2 . 6. Find the following derivatives. Do not simplify. Page 3 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-- 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 . Page 4 Math 171: Exam 2 Solution: Using the quotient rule d 2 y dx 2 = y 2- 2 xy dy dx ( y 2 ) 2 . To put the answer in terms of x and y , we substitute the original expression for dy dx , so d 2 y dx 2 = y 2- 2 xy ( x/y 2 ) y 4 . Page 5...
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2 h h(2 x h 1(2 x 1 = lim h 2(2 x h 1(2 x 1 = 2(2 x 0 1(2 x...

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