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Question
Find \[\frac{dy}{dx}\] when \[x = \frac{2 t}{1 + t^2} \text{ and } y = \frac{1 - t^2}{1 + t^2}\] ?
Solution
\[\text{ We have }, x = \frac{2t}{1 + t^2}\]
\[\Rightarrow \frac{dx}{dt} = \left[ \frac{\left( 1 + t^2 \right)\frac{d}{dt}\left( 2t \right) - 2t\frac{d}{dt}\left( 1 + t^2 \right)}{\left( 1 + t^2 \right)^2} \right] ........\left[ \text{ using quotient rule } \right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{\left( 1 + t^2 \right)\left( 2 \right) - 2t\left( 2t \right)}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{2 + 2 t^2 - 4 t^2}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{2 - 2 t^2}{\left( 1 + t^2 \right)^2} \right] . . . \left( i \right)\]
\[\text{ and,} \]
\[y = \frac{1 - t^2}{1 + t^2}\]
\[\Rightarrow \frac{dy}{dt} = \left[ \frac{\left( 1 + t^2 \right)\frac{d}{dt}\left( 1 - t^2 \right) - \left( 1 - t^2 \right)\frac{d}{dt}\left( 1 + t^2 \right)}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dt} = \left[ \frac{\left( 1 + t^2 \right)\left( - 2t \right) - \left( 1 - t^2 \right)\left( 2t \right)}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dt} = \left[ \frac{- 4t}{\left( 1 + t^2 \right)^2} \right] . . . \left( ii \right)\]
\[\text{ Dividing equation } \left( ii \right) \text{ by } \left( i \right),\text{ we get }, \]
\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{- 4t}{\left( 1 + t^2 \right)^2} \times \frac{\left( 1 + t^2 \right)^2}{2\left( 1 - t^2 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 2t}{1 - t^2}\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{x}{y} .........\left[ \because \frac{x}{y} = \frac{2t}{1 + t^2} \times \frac{1 + t^2}{1 - t^2} = \frac{2t}{1 - t^2} \right]\]
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