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# Hw8webworksolutions - terms y = 2 e 2 x 7 cos x 7 y-1 7 8 Find dy/dx if ln xy = e x y Answer Again I use implicit diﬀerentiation d dx(ln xy = d

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Everybody is assigned slightly diﬀerent versions of each problem. The below are representative examples of the assigned problems, but may not exactly match the problem you were assigned. 1. Use implicit diﬀerentiation to ﬁnd dy/dx , given: e 2 x = sin( x + 7 y ) . Answer: First, diﬀerentiate both sides with respect to x : d dx ( e 2 x ) = d dx (sin( x + 7 y )) e 2 x · 2 = cos( x + 7 y ) · d dx ( x + 7 y ) 2 e 2 x = cos( x + 7 y ) · (1 + 7 y 0 ) by the Chain Rule, or, equivalently, 2 e 2 x = cos( x + 7 y ) + 7 y 0 cos( x + 7 y ) . Isolating the y 0 term yields 7 y 0 cos( x + 7 y ) = 2 e 2 x - cos( x + 7 y ) , so we have y 0 = 2 e 2 x - cos( x + 7 y ) 7 cos( x + 7 y ) . This would be a perfectly acceptable answer, or I could also split the right hand side into two

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Unformatted text preview: terms: y = 2 e 2 x 7 cos( x + 7 y )-1 7 . 8. Find dy/dx if ln( xy ) = e x + y . Answer: Again I use implicit diﬀerentiation: d dx (ln( xy )) = d dx ( e x + y ) 1 xy · d dx ( xy ) = e x + y · d dx ( x + y ) 1 xy ( xy + 1 · y ) = e x + y (1 + y ) xy xy + y xy = e x + y + y e x + y y y + 1 x = e x + y + y e x + y . 1 Grouping the terms containing y yields y ± 1 y-e x + y ² = e x + y-1 x , so we see that y = e x + y-1 x 1 y-e x + y . 2...
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## This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.

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Hw8webworksolutions - terms y = 2 e 2 x 7 cos x 7 y-1 7 8 Find dy/dx if ln xy = e x y Answer Again I use implicit diﬀerentiation d dx(ln xy = d

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