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प्रश्न
Find \[\frac{dy}{dx}\] ,When \[x = e^\theta \left( \theta + \frac{1}{\theta} \right) \text{ and } y = e^{- \theta} \left( \theta - \frac{1}{\theta} \right)\] ?
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उत्तर
\[\text{ We have, x } = e^\theta \left( \theta + \frac{1}{\theta} \right)\]
Differentiating it with respect to \[\theta\]
\[\frac{dx}{d\theta} = e^\theta \frac{d}{d\theta}\left( \theta + \frac{1}{\theta} \right) + \left( \theta + \frac{1}{\theta} \right)\frac{d}{d\theta}\left( e^\theta \right) \left[ \text{ using product rule } \right]\]
\[ \Rightarrow \frac{dx}{d\theta} = e^\theta \left( 1 - \frac{1}{\theta^2} \right) + \left( \frac{\theta^2 + 1}{\theta} \right) e^\theta \]
\[ \Rightarrow \frac{dx}{d\theta} = e^\theta \left( 1 - \frac{1}{\theta^2} + \frac{\theta^2 + 1}{\theta} \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = e^\theta \left( \frac{\theta^2 - 1 + \theta^3 + \theta}{\theta^2} \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = \frac{e^\theta \left( \theta^3 + \theta^2 + \theta - 1 \right)}{\theta^2} . . . \left( i \right)\]
\[\text{ and }, \]
\[ y = e^\theta \left( \theta - \frac{1}{\theta} \right)\]
Differentiating it with respect to \[\theta\] using chain rule
\[\frac{dy}{d\theta} = e^{- \theta} \frac{d}{d\theta}\left( \theta - \frac{1}{\theta} \right) + \left( \theta - \frac{1}{\theta} \right)\frac{d}{d\theta}\left( e^{- \theta} \right) \left[ \text{ using product rule } \right]\]
\[ \Rightarrow \frac{dy}{d\theta} = e^{- \theta} \left( 1 + \frac{1}{\theta^2} \right) + \left( \theta - \frac{1}{\theta} \right) e^\theta \frac{d}{d\theta}\left( - \theta \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = e^{- \theta} \left( 1 + \frac{1}{\theta^2} \right) + \left( \theta - \frac{1}{\theta} \right) e^{- \theta} \left( - 1 \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = e^{- \theta} \left( 1 + \frac{1}{\theta^2} - \theta + \frac{1}{\theta} \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = e^{- \theta} \left( \frac{\theta^2 + 1 - \theta^3 + \theta}{\theta^2} \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = \frac{e^{- \theta} \left( - \theta^3 + \theta^2 + \theta + 1 \right)}{\theta^2} . . . \left( ii \right)\]
\[\text{ Dividing equation } \left( ii \right) by \left( i \right), \]
\[\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = e^{- \theta} \left( \frac{\theta^2 - \theta^3 + \theta + 1}{\theta^2} \right) \times \frac{\theta^2}{e^\theta \left( \theta^3 + \theta^2 + \theta - 1 \right)}\]
\[ = e^{- 2\theta} \left( \frac{\theta^2 - \theta^3 + \theta + 1}{\theta^3 + \theta^2 + \theta - 1} \right)\]
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