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Answer 1

${ \bold{ \underline{Given} : - }} \\ \\ = > x = t + \frac{1}{t} \: \: and \: \: y = t - \frac{1}{t} \\ \\ { \bold{ \underline{To \: \: find} : - }} \\ \\ \frac{ {d}^{2} y}{dx ^{2} } \\ \\ { \bold{ \underline{solution} : - }} \\ \\ \: \: \: \: \: . \: \: Differentiate \: \: 'y' \: \: with \: \: respect \: \: to \: \: 't' - \\ \\ = > \frac{dy}{dt} = 1 + \frac{1}{ {t}^{2} } \: \: \: \: \: \: - - - - - - (1)\\ \\ \: \: \: \: \: . \: \: Differentiate \: \: 'x' \: \: with \: \: respect \: \: to \: \: 't' - \\ \\ = > \frac{dx}{dt} = 1 - \frac{1}{ {t}^{2} } \: \: \: \: \: \: - - - - - - (2) \\ \\ \: \: \: \: \: . \: \: Now \: \: divide \: \: both \: \: equation \: \: (1) \: and \: (2) - \\ \\ = > \frac{dy}{dx} = \frac{1 + \frac{1}{ {t}^{2} } }{1 - \frac{1}{ {t}^{2} } } \\ \\ = > \frac{dy}{dx} = \frac{ {t}^{2} + 1}{ {t}^{2} - 1 } \\ \\ \: \: \: \: \: \: . \: \: Now \: \: Differentiate \: \: \: \: with \: \: respect \: \: to \: \: 'x' - \\ \\ = > \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{( {t}^{2} - 1)(2t) - ( {t}^{2} + 1)(2t)}{ {( {t}^{2} - 1)}^{2} } \times \frac{dt}{dx} \\ \\ = > \frac{ {d}^{2} y}{ {dx}^{2} } = \frac{2t( - 2)}{ { {(t}^{2} - 1)}^{2} } \times \frac{dt}{dx} \\ \\ \: \: \: \: \: \: \: . \: \: by \: \: equation \: (2) - \\ \\ = > \: \frac{dt}{dx} = \frac{ {t}^{2} }{ {t}^{2} - 1 } \\ \\ \: \: \: \: . \: \: so \: \: that \: - \\ \\ = > \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ - 4t}{ {( {t}^{2} - 1) }^{2} } \times \frac{ {t}^{2} }{( {t}^{2} - 1) } \\ \\ = > \frac{ {d}^{2}y }{ {dx}^{2} } = - 4 {t}^{3} {( {t}^{2} -1 ) }^{ - 3} \\ \\ { \bold{ \underline{ Hence \: \: , \: \: option \: \: (b) \: \: is \: \: correct.}}}$