# Exam3_2 3 - x ) + 13 cos(ln x ) cor-rect 2. f ( x ) = 13...

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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert 3 To determine the value of g 0 ( a ), therefore, we need to know b so that f ( b ) = a , for then g 0 ( a ) = 1 f 0 ( b ) . In the given example, f ( x ) = x 3 + 4 x - 3 , a = 13 . Then f 0 ( x ) = 3 x 2 + 4 , f (2) = 13 , in particular, b = 2. Consequently, g 0 (13) = 1 f 0 (2) = 1 16 . keywords: 006 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = e - x (3 sin x - cos x ) . 1. f 0 ( x ) = 4 cos x + 2 sin x 2. f 0 ( x ) = e - x (4 sin x - 2 cos x ) 3. f 0 ( x ) = e - x (sin x - 3 cos x ) 4. f 0 ( x ) = e - x (4 sin x + 2 cos x ) 5. f 0 ( x ) = e - x (3 cos x + sin x ) 6. f 0 ( x ) = e - x (4 cos x - 2 sin x ) correct 7. f 0 ( x ) = 4 cos x - 2 sin x Explanation: By the product rule f 0 ( x ) = - e - x (3 sin x - cos x ) + e - x (3 cos x + sin x ) . Consequently, f 0 ( x ) = e - x (4 cos x - 2 sin x ) . keywords: 007 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = x n 9 sin(ln x ) + 4 cos(ln x ) o . 1. f 0 ( x ) = 5 sin(ln
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Unformatted text preview: x ) + 13 cos(ln x ) cor-rect 2. f ( x ) = 13 sin(ln x ) + 5 cos(ln x ) 3. f ( x ) = 5 sin(ln x )-13 cos(ln x ) 4. f ( x ) = x { 5 sin(ln x ) + 13 cos(ln x ) } 5. f ( x ) = x { 13 sin(ln x ) + 5 cos(ln x ) } Explanation: By the Chain Rule f ( x ) = n 9 sin(ln x ) + 4 cos(ln x ) o + x n 9 cos(ln x ) x-4 sin(ln x ) x o . Consequently, f ( x ) = 5 sin(ln x ) + 13 cos(ln x ) . keywords: 008 (part 1 of 1) 10 points Find the derivative of f ( x ) = 2 sin-1 ( e 3 x ) . 1. f ( x ) = 6 e 3 x 1 + e 6 x...
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