1110-fa2011-PRELIM2-solutions - Math 1110 Prelim 2...

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Math 1110 Prelim 2 (10/27/2011) 1 Question 1. (16 points) Let f ( x ) and g ( x ) be differentiable functions, with f ( x ) a one-to-one func- tion. We know the following values: x f ( x ) g ( x ) f 0 ( x ) g 0 ( x ) - 2 - 2 1 2 8 - 1 1 7 4 1 0 3 9 9 - 3 1 4 - 3 2 6 2 6 - 2 8 - 3 . (a) Let h ( x ) = ln ( x + 2 ) g ( x ) . Then h 0 (- 1 ) = g ( x ) 1 x + 2 - ln ( x + 2 ) g 0 ( x ) ( g ( x )) 2 ± ± ± ± ± x =- 1 = 7 · 1 1 - 0 · 1 7 2 = 1 7 . (b) Let j ( x ) = g ( x ) tan - 1 ( 3x ) . Then j 0 ( 0 ) = ² g 0 ( x ) tan - 1 ( 3x ) + g ( x ) 3 1 + ( 3x ) 2 ³± ± ± ± x = 0 = - 3 · 0 + 9 · 3 1 + 0 = 27. (c) Let ( x ) = f g ( x ) . Then 0 ( 2 ) = f 0 ( g ( x )) g 0 ( x ) ± ± x = 2 = f 0 (- 2 ) · (- 3 ) = 2 · (- 3 ) = - 6. (d) Let p ( x ) = ´ f - 1 ( x ) + 2x 2 µ 4 . The first thing to note is that f (- 1 ) = 1 , so f - 1 ( 1 ) = - 1 . Then p 0 ( 1 ) = 4 h f - 1 ( x ) + 2x 2 i 3 · ² 1 f 0 ( f - 1 ( x )) + 4x ³± ± ± ± x = 1 = 4 · [(- 1 ) + 2 ] 3 · ² 1 f 0 (- 1 ) + 4 ³ = 4 · [ 1 ] 3 · ² 1 4 + 4 ³ = 4 · 17 4 = 17.
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Math 1110 Prelim 2 (10/27/2011) 2 Question 2. (20 points) Evaluate the following expressions. (a) cot ( sin - 1 ( - 1 3 )) e -1 3 Ƽ 8 We first see that sin - 1 (- 1 3 ) is the angle θ given in the figure to the left. Then, taking the cotangent of that angle, we get cot ± sin - 1 ± - 1 3 ²² = 8 - 1 = - 8. (b) d dx h e ln ( x 2 + e ) i = d dx h x 2 + e i = 2x (c) The velocity of a particle is given by s 0 ( t ) = v ( t ) = 3t 4 - 20t 3 + 17t + tan - 1 ( 20 ) . Determine the acceleration a ( t ) . It is a ( t ) = v 0 ( t ) = 12t 3 - 60t 2 + 17 . (d) d dx h x x i for y = x x with x > 0 We use the inverse property of the natural exponential function and the natural log function, then use the chain rule along with the product rule to differentiate.
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This note was uploaded on 03/01/2012 for the course MATH 1110 at Cornell University (Engineering School).

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1110-fa2011-PRELIM2-solutions - Math 1110 Prelim 2...

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