# Solve this

Question:

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

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

Solution:

$\operatorname{as} x=\frac{e^{\theta}+e^{\theta}}{2}$

Differentiating it with respect to $t$

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

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

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{2}\left(\mathrm{e}^{\mathrm{t}}-\mathrm{e}^{-\mathrm{t}}\right)=\mathrm{y}$ ......(1)

And $y=\frac{e^{\theta}-e^{\theta}}{2}$

Differentiating it with respect to $t$,

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

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

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

$\frac{d y}{d t}=\frac{e^{\theta}+e^{\theta}}{2}=X \cdots \cdots(2)$

Dividing equation (2) by (1),

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\mathrm{x}}{\mathrm{y}}$

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