Find y' y^2 = x cos y assuming that the equation determines a differentiable function f such that y= f(x)
2yy' = (1)(cosy) - (siny)y'(x)
2yy'-x(sin(y))y'=cosy
y'=cosy/(2y-xsiny)

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just want to check my answer

2y dy/dx = x ( -sin(y) dy/dx ) +cos(y) )

wait, thats wrong

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