If \(u = x\log xy\), where \({x^3} + {y^3} + 3xy = 1\;\)then \(\frac{{du}}{{dx}}\) is equal to

1
\(\left( {1 + \log xy} \right) - \frac{x}{y}\left( {\frac{{{x^2} + y}}{{{y^2} + x}}} \right)\)
2
\(\left( {1 + \log xy} \right) - \frac{y}{x}\left( {\frac{{{y^2} + x}}{{{x^2} + y}}} \right)\)
3
\(\left( {1 - \log xy} \right) - \frac{x}{y}\left( {\frac{{{x^2} + y}}{{{y^2} + x}}} \right)\)
4
\(\left( {1 - \log xy} \right) - \frac{x}{y}\left( {\frac{{{y^2} + x}}{{{x^2} + y}}} \right)\)

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