sample-midterm2-with-solutions-v2

# Find the third derivative of f x tan x write your

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Unformatted text preview: 2 + 3 y 2 × y ' = 6 y + 6 x × y ' Þ 3 y 2 × y '- 6 x × y ' = 6 y - 3 x 2 Þ 3 y '( y 2 - 2 x) = 3(2 y - x 2 ) 3(2 y - x 2 ) Þ y'= 3( y 2 - 2 x) (2 y - x 2 ) =2 ( y - 2 x) Hence, the slope, m, of the tangent to the given curve at (3,3)is m= 6-9 = -1 9-6 Q9. Find the third derivative of f ( x) = tan( x) . Write your answer in the simplest form. Solution f ¢( x) = sec 2 ( x) f ¢¢( x) = 2 sec( x) · sec( x) · tan( x) = 2 sec 2 ( x) · tan( x) Now we apply product rule, power rule and chain rule: ( f ¢¢¢( x) = 2 · D[sec2 ( x)] · tan( x) + D[tan( x)] · sec 2 ( x) ( = 2 · 2 sec 2 ( x) · tan( x) · tan( x) + sec 2 ( x) · sec 2 ( x) ( ) = 2 sec ( x) · (2 tan ( x) + 1 + tan ( x) ) = 2 sec ( x) · (1 + 3 tan ( x) ) ) ) = 2 sec2 ( x) · 2 tan 2 ( x) + sec 2 ( x) 2 2 2 2 2 4 Q10. A spotlight on the ground shines on a wall 20 m. away. If a man 2 m. tall walks from the spotlight towards the building at a speed of 2.5 m/s, how fast is his shadow on the building decreasing when he is 15 m from the building? Solution Let this length represent the man’s height of 2 m...
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## This document was uploaded on 02/19/2014 for the course MATH Math 113/1 at Grant MacEwan University.

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