Question

Find $\frac{dy}{dx}$:
$x=\frac{e^t+e^{-t}}{2}andy=\frac{e^t-e^{-t}}{2}$

Answer 1

$\begin{array}{c}x=\frac{e^t+e^{-t}}{2}y=\frac{e^t-e^{-t}}{2}\\ \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ now,\\ \frac{dy}{dt}=\frac{e^t+e^{-t}}{2}\\ \frac{dx}{dt}=\frac{e^t-e^{-t}}{2}\\ \frac{dy}{dx}=\frac{e^t+e^{-t}}{e^t-e^{-t}}=\frac{x}{y}\end{array}$