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

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


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

$\Rightarrow \frac{d x}{d t}=\frac{1}{2}\left[\frac{d}{d t}\left(e^{t}\right)+\frac{d}{d t}\left(e^{-t}\right)\right]$ and $\frac{d y}{d t}=\frac{1}{2}\left[\frac{d}{d t}\left(e^{t}\right)-\frac{d}{d t} e^{-t}\right]$

$\Rightarrow \frac{d x}{d t}=\frac{1}{2}\left[e^{t}+e^{-t} \frac{d}{d t}(-t)\right]$ and $\frac{d y}{d t}$

$=\frac{1}{2}\left[e^{t}-e^{-t} \frac{d}{d t}\left(e^{-t}\right)\right]$

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

$=y$ and $\frac{d y}{d t}=\frac{1}{2}\left(e^{t}+e^{-t}\right)=x$

$\therefore \frac{d y}{d t}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{x}{y}$

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