What is the implicit derivative of #1=x/y-cosy#?

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Answer 1 · 2016-09-02T15:41:48

#dy/dx = -y/(y^2siny - x)#

#d/dx(1) = d/dx(x/y - cosy)#

#d/dx(1) = d/dx(x/y) + d/dx(-cosy)#

#0 = (1 xx y - 1 xx x(dy/dx))/y^2 + siny(dy/dx)#

#-siny(dy/dx) = (y - x(dy/dx))/y^2#

#-siny(dy/dx)y^2 = y - x(dy/dx)#

#-siny(dy/dx)y^2 + x(dy/dx) = y#

#dy/dx(-y^2siny + x) = y#

#dy/dx = y/(-y^2siny + x)#

#dy/dx = -y/(y^2siny - x)#

Hopefully this helps!