Question

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

$x=\frac{e^t+e^{-t}}{2}\mathrm{and}y=\frac{e^t-e^{-t}}{2}$

Answer 1

$\mathrm{We}\mathrm{have},x=\frac{e^t+e^{-t}}{2}\mathrm{and}y=\frac{e^t-e^{-t}}{2}$

$$\begin{gathered} \Rightarrow \frac{dx}{dt}=\frac{1}{2}\left[\frac{d}{dt}\left(e^t\right)+\frac{d}{dt}\left(e^{-t}\right)\right]\mathrm{and}\frac{dy}{dt}=\frac{1}{2}\left[\frac{d}{dt}\left(e^t\right)-\frac{d}{dt}{e}^{-t}\right] \\ \Rightarrow \frac{dx}{dt}=\frac{1}{2}\left[e^t+e^{-t}\frac{d}{dt}\left(-t\right)\right]\mathrm{and}\frac{dy}{dt}=\frac{1}{2}\left[e^t-e^{-t}\frac{d}{dt}\left(e^{-t}\right)\right] \\ \Rightarrow \frac{dx}{dt}=\frac{1}{2}\left(e^t-e^{-t}\right)=y\mathrm{and}\frac{dy}{dt}=\frac{1}{2}\left(e^t+e^{-t}\right)=x \\ \therefore \frac{dy}{dt}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{x}{y} \end{gathered}$$