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(D) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = xy$.Separating the variables,we get $\frac{1}{y} dy = x dx$.Integratin

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$2 \log y = x^{2} - 1$

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(D) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = xy$.Separating the variables,we get $\frac{1}{y} dy = x dx$.Integrating both sides,we have $\int \frac{1}{y} dy = \int x dx$,which gives $\log y = \frac{x^{2}}{2} + C$.Since the curve passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:$\log(1) = \frac{1^{2}}{2} + C \implies 0 = \frac{1}{2} + C \implies C = -\frac{1}{2}$.Substituting the value of $C$ back into the equation,we get $\log y = \frac{x^{2}}{2} - \frac{1}{2}$.Multiplying by $2$,we obtain $2 \log y = x^{2} - 1$.

The equation of the curve passing through the point $(1, 1)$ such that the slope of the tangent at any point $(x, y)$ is equal to the product of its coordinates is

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