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Question 629779: find dy/dx for 2sinxcosy=1
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! Using the product rule, (uv)' = u*v' + v*u', with u = 2cos(x) and v = sin(y)
2sin(x)*(-sin(y)*y') + cos(y)*(2cos(x)) = 0
Simplifying:
-2sin(x)sin(y)*y' + 2cos(x)cos(y) = 0
Solving for y':
-2sin(x)sin(y)*y' = - 2cos(x)cos(y)
y' = (-2cos(x)cos(y))/(-2sin(x)sin(y))
which reduces to:
y' = (cos(x)cos(y))/(sin(x)sin(y))
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