Solve this

Question:

Find $\frac{d y}{d x}$, when

$x=e^{\theta}\left(\theta+\frac{1}{\theta}\right)$ and $y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

Solution:

We have, $x=e^{\theta}\left(\theta+\frac{1}{\theta}\right)$

Differentiating it with respect to $\theta$,

$\frac{d x}{d \theta}=e^{\theta} \frac{d}{d \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \frac{d}{d \theta}\left(e^{\theta}\right)$             [using product rule]

$\Rightarrow \frac{d x}{d \theta}=e^{\theta}\left(1-\frac{1}{\theta^{2}}\right)+\left(\frac{\theta^{2}+1}{\theta}\right) e^{\theta}$

$\Rightarrow \frac{d x}{d \theta}=e^{\theta}\left(1-\frac{1}{\theta^{2}}+\frac{\theta^{2}+1}{\theta}\right)$

$\Rightarrow \frac{d x}{d \theta}=e^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)$

$\Rightarrow \frac{d x}{d \theta}=\frac{e^{\theta}\left(\theta^{3}+\theta^{2}+\theta-1\right)}{\theta^{2}}$                .....(1)

and,

$y=e^{\theta}\left(\theta-\frac{1}{\theta}\right)$

Differentiating it with respect to $\theta$ using chain rule,

$\frac{d y}{d \theta}=e^{-\theta} \frac{d}{d \theta}\left(\theta-\frac{1}{\theta}\right)+\left(\theta-\frac{1}{\theta}\right) \frac{d}{d \theta}\left(e^{-\theta}\right)$                [using product rule]

$\Rightarrow \frac{d y}{d \theta}=e^{-\theta}\left(1+\frac{1}{\theta^{2}}\right)+\left(\theta-\frac{1}{\theta}\right) e^{\theta} \frac{d}{d \theta}(-\theta)$

$\Rightarrow \frac{d y}{d \theta}=e^{-\theta}\left(1+\frac{1}{\theta^{2}}\right)+\left(\theta-\frac{1}{\theta}\right) e^{-\theta}(-1)$

$\Rightarrow \frac{d y}{d \theta}=e^{-\theta}\left(1+\frac{1}{\theta^{2}}-\theta+\frac{1}{\theta}\right)$

$\Rightarrow \frac{d y}{d \theta}=e^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)$

$\Rightarrow \frac{d y}{d \theta}=\frac{e^{-\theta}\left(-\theta^{3}+\theta^{2}+\theta+1\right)}{\theta^{2}}$         .....(2)

Dividing equation (ii) by (i),

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=e^{-\theta}\left(\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{2}}\right) \times \frac{\theta^{2}}{e^{\theta}\left(\theta^{3}+\theta^{2}+\theta-1\right)}$

$=e^{-2 \theta}\left(\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right)$

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