sol13-2 - y = e x ln( x ) (1 ln( x ) + x 1 x ) 32. Find the...

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SOLUTIONS CHAPTER 13.2 MATH 132 WI01 4. y 0 = e 2 x 2 +5 · (2 · 2 x ) = e 2 x 2 +5 · 4 x 14. y = e x - e - x 2 = e x 2 - e - x 2 = 1 2 e x - 1 2 e - x (you will see in a minute why I prefer to keep the negative power as it is and not change it into a positive power in the denominator, as in e - x = 1 e x ) y 0 = 1 2 e x - 1 2 e - x · ( - 1) (for the last term use chain rule!); simplified (common factor the half, and the two minuses that will give you a plus) y 0 = 1 2 ( e x + e - x ) 26. Product rule: y 0 = e - x ( - 1) · ln( x ) + e - x · 1 x 27. Chain rule, followed by product rule (for the power!)
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Unformatted text preview: y = e x ln( x ) (1 ln( x ) + x 1 x ) 32. Find the derivative of y rst (use product rule): y = ex e-1 e x + x e e x Now plug in 1 for x and produce the slope= e 1 e-1 e 1 + 1 e e 1 = e 2 + e and now use point-slope formula : y-e = ( e 2 + e )( x-1) Date : 02/07/2000. 1...
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This note was uploaded on 07/26/2011 for the course MATH 132 taught by Professor Staff during the Spring '08 term at Ohio State.

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