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

May 7, 2016

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

#### Explanation:

When equation are given in parametric form such as

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

Then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

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

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1 \times \sqrt{{t}^{2} - 1} + t \times \frac{1}{2} \times \frac{1}{\sqrt{{t}^{2} - 1}}$

= $\sqrt{{t}^{2} - 1} + \frac{t}{2} \times \frac{1}{\sqrt{{t}^{2} - 1}}$

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

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