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the derivative of x is 1
the derivative of 2 is 0
and for the derivative of \(xy\) you need the product rule \[xy'+y\]
so you get \[xy'+y=1\] solve for \(y'\)
if that is not clear, imagine \(y=f(x)\) you take the derivative of \(xf(x)\) by the product rule and get \(xf'(x)+f(x)\)
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yes \(\frac{1-y}{x}\) from algebra
In this case, if you were desperate, you could first solve for y. Not really the point of the exercise, though.
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