Exam2_2 2 - x + 5)-1 / 2 , while f 00 ( x ) =-1 2 4 2(4 x +...

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Taylor, Douglas – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: JEGilbert 2 6. f 00 ( x ) = 14cos2 x Explanation: Diferentiating once we see that f 0 ( x ) = - 6sin2 x - 2sin x cos x . Now 2sin x cos x = sin2 x , so f 0 ( x ) = - 7sin2 x . Consequently, by diferentiating once again we obtain f 00 ( x ) = - 14cos2 x . keywords: second derivative, trig Function 004 (part 1 oF 1) 10 points Determine the third derivative, f 000 ( x ), oF f when f ( x ) = 4 x + 5 . 1. f 000 ( x ) = - 24(4 x + 5) - 3 / 2 2. f 000 ( x ) = - 4(4 x + 5) - 3 / 2 3. f 000 ( x ) = 24(4 x + 5) - 5 / 2 correct 4. f 000 ( x ) = 4(4 x + 5) - 3 / 2 5. f 000 ( x ) = 4(4 x + 5) - 5 / 2 6. f 000 ( x ) = - 24(4 x + 5) - 5 / 2 Explanation: To use the Chain Rule successively it’s more convenient to write f ( x ) = 4 x + 5 = (4 x + 5) 1 / 2 . ±or then f 0 ( x ) = 1 2 · 4 · (4 x + 5) - 1 / 2 = 2(4
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Unformatted text preview: x + 5)-1 / 2 , while f 00 ( x ) =-1 2 4 2(4 x + 5)-3 / 2 =-4(4 x + 5)-3 / 2 , and f 000 ( x ) = 3 2 4 4(4 x + 5)-5 / 2 . Consequently, f 000 ( x ) = 24(4 x + 5)-5 / 2 . keywords: third derivative, chain rule 005 (part 1 oF 1) 10 points ind dy/dx when 2 x 2 + 3 y 2 = 6 . 1. dy dx = 3 xy 2. dy dx =-2 x y 3. dy dx =-2 x 3 y correct 4. dy dx = 2 x 3 y 5. dy dx =-2 xy 6. dy dx = x 3 y Explanation: DiFerentiating 2 x 2 + 3 y 2 = 6 implicitly with respect to x we see that 4 x + 6 y dy dx = 0 ....
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