dy/dx of xy^2=tan(2x+4y)
Question ID
539598
Created
April 30, 2011 6:30pm UTC
Rating
0
URL
https://questions.llc/questions/539598
Answers
3
Views
247
3 answers

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
539600Created
April 30, 2011 6:32pm UTCRating
0 
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
539622Created
April 30, 2011 6:57pm UTCRating
0 
THanks Reiny :)
Answer ID
539624Created
April 30, 2011 6:59pm UTCRating
0