Question

Find $\frac{dy}{dx}$, when

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

Answer 1

$\mathrm{We}\mathrm{have},x=e^{\theta }\left(\theta +\frac{1}{\theta }\right)$
Differentiating it with respect to $\theta$,
$$\begin{gathered} \frac{dx}{d\theta }=e^{\theta }\frac{d}{d\theta }\left(\theta +\frac{1}{\theta }\right)+\left(\theta +\frac{1}{\theta }\right)\frac{d}{d\theta }\left(e^{\theta }\right)\left[\mathrm{using}\mathrm{product}\mathrm{rule}\right] \\ \Rightarrow \frac{dx}{d\theta }=e^{\theta }\left(1-\frac{1}{{\theta }^2}\right)+\left(\frac{{\theta }^2+1}{\theta }\right)e^{\theta } \\ \Rightarrow \frac{dx}{d\theta }=e^{\theta }\left(1-\frac{1}{{\theta }^2}+\frac{{\theta }^2+1}{\theta }\right) \\ \Rightarrow \frac{dx}{d\theta }=e^{\theta }\left(\frac{{\theta }^2-1+{\theta }^3+\theta }{{\theta }^2}\right) \\ \Rightarrow \frac{dx}{d\theta }=\frac{e^{\theta }\left({\theta }^3+{\theta }^2+\theta -1\right)}{{\theta }^2}...\left(i\right) \\ \mathrm{and}, \\ y=e^{\theta }\left(\theta -\frac{1}{\theta }\right) \end{gathered}$$
Differentiating it with respect to $\theta$ using chain rule,

$$\begin{gathered} \frac{dy}{d\theta }=e^{-\theta }\frac{d}{d\theta }\left(\theta -\frac{1}{\theta }\right)+\left(\theta -\frac{1}{\theta }\right)\frac{d}{d\theta }\left(e^{-\theta }\right)\left[\mathrm{using}\mathrm{product}\mathrm{rule}\right] \\ \Rightarrow \frac{dy}{d\theta }=e^{-\theta }\left(1+\frac{1}{{\theta }^2}\right)+\left(\theta -\frac{1}{\theta }\right)e^{\theta }\frac{d}{d\theta }\left(-\theta \right) \\ \Rightarrow \frac{dy}{d\theta }=e^{-\theta }\left(1+\frac{1}{{\theta }^2}\right)+\left(\theta -\frac{1}{\theta }\right)e^{-\theta }\left(-1\right) \\ \Rightarrow \frac{dy}{d\theta }=e^{-\theta }\left(1+\frac{1}{{\theta }^2}-\theta +\frac{1}{\theta }\right) \\ \Rightarrow \frac{dy}{d\theta }=e^{-\theta }\left(\frac{{\theta }^2+1-{\theta }^3+\theta }{{\theta }^2}\right) \\ \Rightarrow \frac{dy}{d\theta }=\frac{e^{-\theta }\left(-{\theta }^3+{\theta }^2+\theta +1\right)}{{\theta }^2}...\left(ii\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\left(\mathrm{i}\right), \\ \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=e^{-\theta }\left(\frac{{\theta }^2-{\theta }^3+\theta +1}{{\theta }^2}\right)\times \frac{{\theta }^2}{e^{\theta }\left({\theta }^3+{\theta }^2+\theta -1\right)} \\ =e^{-2\theta }\left(\frac{{\theta }^2-{\theta }^3+\theta +1}{{\theta }^3+{\theta }^2+\theta -1}\right) \end{gathered}$$