Asked by philip
y=cosxcosy-sinxsiny/sinxcosy+cosxsiny
find dy/dx
plz show me working
find dy/dx
plz show me working
Answers
Answered by
Reiny
I assume you meant:
y=(cosxcosy-sinxsiny)/(sinxcosy+cosxsiny)
= cos(x+y) / sin(x+y) , using your identities
= cot(x+y)
dy/dx = - csc^2 (x+y) (1 + dy/dx)
dy/dx = -csc^2 (x+y) - dy/dx csc^2 (x+y)
dy/dx + dy/dx csc^2 (x+y) = -csc^2 (x+y)
dy/dx(1 + csc^2 (x+y) ) = -csc^2 (x+y)
dy/dx = -csc^2 (x+y)/(1 + csc^2 (x+y) )
y=(cosxcosy-sinxsiny)/(sinxcosy+cosxsiny)
= cos(x+y) / sin(x+y) , using your identities
= cot(x+y)
dy/dx = - csc^2 (x+y) (1 + dy/dx)
dy/dx = -csc^2 (x+y) - dy/dx csc^2 (x+y)
dy/dx + dy/dx csc^2 (x+y) = -csc^2 (x+y)
dy/dx(1 + csc^2 (x+y) ) = -csc^2 (x+y)
dy/dx = -csc^2 (x+y)/(1 + csc^2 (x+y) )
Answered by
Steve
and, since y=cot(x+y),
dy/dx = -(y^2+1)/(y^2+2)
dy/dx = -(y^2+1)/(y^2+2)
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