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Mathematics 14 Online
OpenStudy (chris215):

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OpenStudy (anonymous):

the derivative of x is 1

OpenStudy (anonymous):

the derivative of 2 is 0

OpenStudy (anonymous):

and for the derivative of \(xy\) you need the product rule \[xy'+y\]

OpenStudy (anonymous):

so you get \[xy'+y=1\] solve for \(y'\)

OpenStudy (anonymous):

if that is not clear, imagine \(y=f(x)\) you take the derivative of \(xf(x)\) by the product rule and get \(xf'(x)+f(x)\)

OpenStudy (anonymous):

yes \(\frac{1-y}{x}\) from algebra

OpenStudy (tkhunny):

In this case, if you were desperate, you could first solve for y. Not really the point of the exercise, though.

OpenStudy (chris215):

Thank you!!!

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