If u = x log xy, where x3 + y3 + 3xy = 1, then \(\frac{{du}}{{dx}}\) is

\(1 + \log xy + \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)

\(1 - \log xy + \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)

\(1 - \log xy - \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)

\(1 + \log xy - \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)

If u = x log xy, where x3 + y3 + 3xy = 1, then \(\frac{{du}}{{dx}}\)

1.	If u = x log xy, where x3 + y3 + 3xy = 1, then \(\frac{{du}}{{dx}}\) is
A.	\(1 - \log xy + \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)
B.	\(1 - \log xy - \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)
C.	\(1 + \log xy - \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)
D.	\(1 + \log xy + \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)
Answer» D. \(1 + \log xy + \frac{{x\left( {{x^2} + y} \right)}}{{y\;\left( {{y^2} + x} \right)}}\)

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If u = x log xy, where x3 + y3 + 3xy = 1, then \(\frac{{du}}{{dx}}\) is

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