# How do you differentiate the following parametric equation:  x(t)=t/sqrt(t^2-1), y(t)= sqrt(t^2-e^(t) ?

Oct 22, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({e}^{t} - 2 t\right) \left({\sqrt{{t}^{2} - 1}}^{3}\right)}{2 \sqrt{{t}^{2} - {e}^{t}}}$

#### Explanation:

I assume that you want to find $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Given the parametric equations $x \left(t\right)$ and $y \left(t\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$.

We are given $y \left(t\right) = \sqrt{{t}^{2} - {e}^{t}}$. Using the chain rule, $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2 t - {e}^{t}}{2 \sqrt{{t}^{2} - {e}^{t}}}$.

Using $x \left(t\right) = \frac{t}{\sqrt{{t}^{2} - 1}}$ and the quotient rule, $\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\sqrt{{t}^{2} - 1} - {t}^{2} / \sqrt{{t}^{2} - 1}}{{t}^{2} - 1} = - \frac{1}{\sqrt{{t}^{2} - 1}} ^ 3$.

Thus, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{\frac{2 t - {e}^{t}}{2 \sqrt{{t}^{2} - {e}^{t}}}}{- \frac{1}{\sqrt{{t}^{2} - 1}} ^ 3} = \frac{2 t - {e}^{t}}{2 \sqrt{{t}^{2} - {e}^{t}}} \cdot - {\sqrt{{t}^{2} - 1}}^{3} = \frac{\left({e}^{t} - 2 t\right) \left({\sqrt{{t}^{2} - 1}}^{3}\right)}{2 \sqrt{{t}^{2} - {e}^{t}}}$