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

Mar 29, 2016

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

#### Explanation:

For the parametric function the derivative is given by

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

rewriting x(t) as $x \left(t\right) = t {\left({t}^{2} - 1\right)}^{\frac{1}{2}} \text{ for ease of differentiating }$

Now have a product of 2 functions, which can be differentiated using the $\textcolor{b l u e}{\text{ product rule }}$

If f(x) = g(x).h(x) then f'(x) = g(x).h'(x) + h(x).g'(x)
$\text{------------------------------------------------------------}$
so x'(t) = t . $\frac{d}{\mathrm{dt}} {\left({t}^{2} - 1\right)}^{\frac{1}{2}} + {\left({t}^{2} - 1\right)}^{\frac{1}{2}} . \frac{d}{\mathrm{dt}} \left(t\right)$

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

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

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

rewriting as a single fraction.

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

and $y ' \left(t\right) = 2 t - {e}^{t}$
$\text{------------------------------------------------}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} = \frac{2 t - {e}^{t}}{\frac{2 {t}^{2} - 1}{{t}^{2} - 1} ^ \left(\frac{1}{2}\right)}$

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