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# Midterm - Midterm 2 SOLUTIONS MAT 131 Fall 2011 1 Calculate...

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Midterm 2 SOLUTIONS MAT 131 Fall 2011 1. Calculate the derivatives of the following functions: (a) f ( x ) = 3 x 3 + 4 x 2 + 5 x + 6 x f 0 ( x ) = 3(3 x 2 ) + 4(2 x ) + 5 x 0 + 6( - 1) x - 2 = 9 x 2 + 8 x + 5 - 6 x 2 (b) f ( x ) = sin 10 x f 0 ( x ) = 10(sin x ) 9 d dx sin x = 10 sin 9 x cos x (c) f ( x ) = x 2 + 1 x + 5 f 0 ( x ) = ( x + 5)(2 x ) - ( x 2 + 1)(1) ( x + 5) 2 = 2 x 2 + 10 x - x 2 - 1 ( x + 5) 2 = x 2 + 10 x - 1 ( x + 5) 2 (d) f ( x ) = 3 x log 3 ( x + 3) f 0 ( x ) = 3 x d dx ln( x + 3) ln 3 + log 3 ( x + 3) d dx 3 x = 3 x 1 (ln 3)( x + 3) + 3 x (ln 3) log 3 ( x + 3) = 3 x (ln 3)( x + 3) + 3 x ln( x + 3) (e) f ( x ) = arctan( x 2 + 1) f 0 ( x ) = 1 1 + ( x 2 + 1) 2 · 2 x = 2 x x 4 + 2 x 2 + 2 (f) f ( x ) = x sin x First we set y = f ( x ) and take a logarithm: ln y = ln( x sin x ) = sin x ln x . Next we differentiate both sides implicitly with respect to x : y 0 y = sin x x + cos x ln x. Then we multiply both sides by y and replace y with f ( x ): f 0 ( x ) = x sin x sin x x + cos x ln x

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MAT 131 Fall 2011 Midterm 2 SOLUTIONS Page 2 2. (a) Find a linear function that best approximates tan x at the point x = π 4 . We need to know the values of tan x and (tan x ) 0 when x = π/ 4. Because (tan x ) 0 = sec 2 x , we compute tan π 4 = 1 , sec 2 π 4 = 1 cos 2 ( π/ 4) = 1 ( 2 / 2) 2 = 2 .
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