124HW18solutions

# 124HW18solutions - 61 We ﬁrst diﬀerentiate f x = 3 x-1 2 Hence the slope of the tangent line at x = 2 is m = f(2 = 3(1 2 = 3 Noting that f(2

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Math 124 Homework 18 (section 3.4), solutions to graded problems 20) By two applications of the Chain Rule, we have g 0 ( t ) = e (1+3 t ) 2 2(1 + 3 t )3 = 6(1 + 3 t ) e (1+3 t ) 2 29) w 0 = 3 2 ( x 2 · 5 x ) 1 2 ( x 2 (ln 5)5 x + 2 x 5 x ) 51) f 0 ( x ) = a ( x ( - b ) e - bx + e - bx ) = - abxe - bx + ae - bx 54) This involves a “symbolic” use of the Chain Rule: u 0 ( x ) = g 0 ( f ( x )) f 0 ( x ) (a) u 0 (1) = g 0 ( f (1)) f 0 (1) = g 0 (2)(2) = - 1(2) = - 2 (b) u 0 (2) = g 0 ( f (2)) f 0 (2) = undeﬁned , since f 0 (2) is undeﬁned.
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Unformatted text preview: 61) We ﬁrst diﬀerentiate: f ( x ) = 3( x-1) 2 . Hence the slope of the tangent line at x = 2 is m = f (2) = 3(1) 2 = 3. Noting that f (2) = (2-1) 3 = 1 and using the point-slope form for a line, we have y-1 = 3( x-2) or y = 3 x-5 1...
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## This note was uploaded on 06/16/2008 for the course MATH 124 taught by Professor Kennedy during the Spring '08 term at University of Arizona- Tucson.

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