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sol13-1 - 1 2 x 3 2 42 Use ln’s properties-ln x 2 √...

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SOLUTIONS CHAPTER 13.1 MATH 132 WI01 8. y 0 = 1 - x 2 + 6 x · ( - 2 x + 6) 14. Product rule + chain rule: [2(2 x + 5) · 2] · ln(2 x + 5) + (2 x + 5) 2 · [ 1 2 x + 5 · 2] 18. Use the fact that log 4 ( x ) = ln( x ) ln(4) and rewrite the function: y = x 3 log 4 ( x ) = x 3 ln( x ) ln(4) = 1 ln(4) · x 3 ln( x ) (where, of course, ln(4) is a constant , right?). Hence y 0 = 1 ln(4) (3 x 2 ln( x ) + x 3 1 x ) 40. What we have here is a power which is not inside the ln, but outside (formally: applied to ln), so we have to use chain rule first of all: y 0 = 2 ln(2 x
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Unformatted text preview: ( 1 2 x + 3 · 2) 42. Use ln’s properties -ln( x 2 · √ 3 x-2) = ln( x 2 ) + ln( √ 3 x-2) , ln( x 2 ) = 2ln( x ) and ln( √ 3 x-2) = ln((3 x-2) 1 2 ) = 1 2 ln(3 x-2) . Rewrite the function: y = ln( x 2 √ 3 x-2) = ln( x 2 ) + ln( √ 3 x-2) = 2ln( x ) + 1 2 ln(3 x-2) Date : 02/07/2000. 1 2 MATH 132 WI01 and so y = 2 1 x + 1 2 1 3 x-2 · 3...
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