Question

x = 1/(1 - t ^ 2), y = 1 + t ^ 2 ,,,then find derivative of. dy/dx
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Answer 1

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$\huge\underline\mathcal\red{Answer}$

$\underline{\underline{{\purple{\sf Given\: values}}}}$

$\: \: \: \: \: \star \bf{x = \frac{1}{(1 - {t}^{2} )} \: \: and \: y = 1 + {t}^{2} }$

$\: \: \: \: \: \: \bigstar \boxed{ \rm{ \: to \: find \: the \: derivative \: of \frac{dy}{dx} }}$

$\: \: \: \: \: \ \ \dashrightarrow \boxed{ \sf \pink{solution}}$

$\: \: \: \: \: \: \implies \rm{x = \frac{1}{(1 - {t}^{2} )} }$

$\: \: \: \: \clubsuit \bf {differentiating \: equation \: with \: respect \: to \: t}$

$\: \: \: \: \blue \star \boxed{ \sf{x = {(1 - {t}^{2} )}^{ - 1} }}$

$\: \: \: \: \implies \sf{ \frac{dx}{dt} = - 1(1 - {t}^{2} )^{ - 2} \times (0 - 2t)}..........by \: using \: chain \: rule$

$\: \: \: \implies \sf{ \frac{dx}{dt }= ( - 1)( - 2t)(1 - {t}^{2} )}$

$\: \: \: \: \implies \sf{ \frac{dx}{dt } = 2t(1 - {t}^{2} ).......(1)eqn}$

$\: \: \: \: \blue \star \boxed{ \sf{y = 1 + {t}^{2} )}}$

$\: \implies \bf{ differentiating \: with \: respect \: to \: t}$

$\: \: \: \: \implies \sf{ \frac{dy}{dt} = 1 + {t}^{2} }$

$\: \: \: \: \implies \sf{ \frac{dy}{dt} = 0 + 2t}$

$\: \: \: \: \implies \sf{ \: \frac{dy}{dt} = 2t} \: \: \: \: \: \: \: ...........(2)eqn$

$\: \: \: \: \boxed{ \bf{derivative \: of \: \frac{dy}{dx} is}}$

$\: \: \: \: \: \: \: \boxed{ \rm{ \: dividing \: both \: eqn \: to \: find \: derivative \: of \: \frac{dy}{dx} }}$

$\: \: \: \: \: \implies \sf{ \frac{dy}{dx} = \frac{2t}{2t(1 - {t}^{2} )} }$

$\: \: \: \: \implies \sf{ \frac{dy}{dx} = \cancel \frac{2t}{2t} \times \frac{1}{(1 - {t}^{2}) } }$

$\: \: \bigstar \sf\frac{dy}{dx} = \frac{1}{(1 - {t}^{2}) }$

$\: \: \: \: \dashrightarrow \sf{the \: derivative \: of \: \frac{dy}{dx} is \: \frac{1}{(1 - {t}^{2}) } }$

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