dy/dx of xy^2=tan(2x+4y)

3 answers

1. i got the answer up to
   2xy dy/dx+y^2=sec^2(2x+4y)(2+4 dy/dx) but i can figure out the rest :(

   Answer ID
   539600

   Created
   April 30, 2011 6:32pm UTC

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   URL
   https://questions.llc/answers/539600

2. Your derivative so fare is correct, now
   expand your right side to ..
   
   2sec^2(2x+4y) + 4 dy/dx sec^2(2x+4y)
   
   bring both terms containing dy/dx to one side, the other two terms to the other side.
   Factor out dy/dx, then isolated dy/dx by dividing by the remaining factor.
   
   I see it as
   dy/dx = (2sec^2(2x+4y) - y^2)/(2xy - 4sec^2(2x+4y))

   Answer ID
   539622

   Created
   April 30, 2011 6:57pm UTC

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   URL
   https://questions.llc/answers/539622

3. THanks Reiny :)

   Answer ID
   539624

   Created
   April 30, 2011 6:59pm UTC

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   0

   URL
   https://questions.llc/answers/539624