If {x^2} + {y^2} = t - frac{1}{t} and {x^4} + | vedclass

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(B) Given: ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$.We know that ${({x^2} + {y^2})^2} = {x^4} + {y^4} + 2{x^2

Vedclass · Accepted Answer

$1/({x^3}y)$

Vedclass · Answer

(B) Given: ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$.We know that ${({x^2} + {y^2})^2} = {x^4} + {y^4} + 2{x^2}{y^2}$.Substituting the given values: ${\left( {t - \frac{1}{t}} \right)^2} = {t^2} + \frac{1}{{{t^2}}} + 2{x^2}{y^2}$.Expanding the left side: ${t^2} - 2 + \frac{1}{{{t^2}}} = {t^2} + \frac{1}{{{t^2}}} + 2{x^2}{y^2}$.This simplifies to $-2 = 2{x^2}{y^2}$,which means ${x^2}{y^2} = -1$,or ${y^2} = -\frac{1}{{{x^2}}}$.Differentiating both sides with respect to $x$: $2y \frac{{dy}}{{dx}} = -\frac{d}{{dx}}({x^{-2}}) = -(-2){x^{-3}} = \frac{2}{{{x^3}}}$.Therefore,$\frac{{dy}}{{dx}} = \frac{2}{{{x^3} \cdot 2y}} = \frac{1}{{{x^3}y}}$.

If ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$,then $\frac{{dy}}{{dx}}$ equals:

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