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# Find $\large \frac{dy}{dx}$ of the function expressed in parametric form $x=e^\theta\bigg(\theta+\frac{1}{\theta}\bigg),y=e^{-\theta}\bigg(\theta-\frac{1}{\theta}\bigg)$

Toolbox:
• To find $\large\frac{dy}{dx}$ in the case of parametric functions,if $x=\phi(t)$ and $y=\psi(t)$,then $\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}} Step 1: y=e^{-\theta}\big(\theta-\large\frac{1}{\theta}\big) \;\;=e^{-\theta}\theta-\large\frac{-1}{\theta}.e^{-\theta} Now differentiating w.r.t \theta by applying product rule, Consider y=e^{-\theta}(\theta-\large\frac{1}{\theta}) Let u=\theta \large\frac{du}{d\theta}$$=1$
Let $v=e^{-\theta}$
$\large\frac{dv}{d\theta}=-e^{-\theta}$
Therefore $\large\frac{dy}{d\theta}=u.\large\frac{dv}{d\theta}+v.\large\frac{du}{d\theta}$
$\Rightarrow \theta(-e^{-\theta})+e^{-\theta}.1$
$\Rightarrow e^{-\theta}(-\theta+1)$
Let $u=\large\frac{1}{\theta}$
$\large\frac{du}{d\theta}$$=-\large\frac{1}{\theta^2} Let v=e^{-\theta} \large\frac{dv}{d\theta}=-e^{-\theta} \large\frac{dy}{d\theta}=\large\frac{1}{\theta}$$.(-e^{-\theta}).\big(\large\frac{-1}{\theta^2}\big)$
$\qquad=-e^{-\theta}\big[\large\frac{1}{\theta}+\frac{1}{\theta^2}\big]$
Step 2:
As done previously let us differentiate w.r.t $\theta$ by applying product rule,
Let $u=\theta$
$\large\frac{du}{d\theta}$$=1 Let v=e^{\theta} Therefore \large\frac{dx}{d\theta}$$=\theta.e^\theta+e^\theta.1$
$\Rightarrow e^\theta(\theta+1)$
Let $u=\large\frac{1}{\theta}$
$\large\frac{du}{d\theta}$$=-\large\frac{1}{\theta^2} Let v=e^{\theta} \large\frac{dv}{d\theta}=e^{\theta} Therefore \large\frac{dx}{d\theta}$$=\large\frac{1}{\theta}$$.e^\theta+e^\theta.\big(\large\frac{-1}{\theta^2}\big) \qquad\qquad\;\;=e^\theta(\large\frac{1}{\theta}-\large\frac{1}{\theta^2}) \large\frac{dy}{d\theta}=$$e^{-\theta}[1-\theta]-(-e^{-\theta})(\large\frac{1}{\theta}+\frac{1}{\theta^2})$
$\Rightarrow e^{\theta}[1-\theta+\large\frac{1}{\theta}+\frac{1}{\theta^2}]$
$\large\frac{dx}{d\theta}=$$e^{-\theta}[1-\theta]-(-e^{-\theta})(\large\frac{1}{\theta}+\frac{1}{\theta^2}) Step 3: Therefore \large\frac{dy}{dx}$$=\large\frac{e^{-\theta}}{e^\theta}\large\frac{(1-\theta+\large\frac{1}{\theta}+\large\frac{1}{\theta}^2)}{(1+\theta+\large\frac{1}{\theta}-\large\frac{1}{\theta^2})}$
$\qquad\qquad\;\;\;=e^{-2\theta}\bigg[\large\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\bigg]$