# What is the derivative of f(t) = (t^2-1 , te^(2t^2-2) ) ?

Jun 19, 2017

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

#### Explanation:

Here the function is desccribed in parametric form as

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

In such cases $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t$

and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1 \times {e}^{2 {t}^{2} - 2} + t \times {e}^{2 {t}^{2} - 2} \times 4 t$

= ${e}^{2 {t}^{2} - 2} \left(1 + 4 {t}^{2}\right)$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{2 {t}^{2} - 2} \left(1 + 4 {t}^{2}\right)}{2 t}$

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