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

Aug 3, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {e}^{2 {t}^{2} - 2}$

#### Explanation:

In parametric form of function where $f \left(t\right) = \left(g \left(t\right) , h \left(t\right)\right)$

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

as such derivative of given function is

(d/dt(e^(2t^2-2)))/(d/(dt)(t^2+1)

= $\frac{{e}^{2 {t}^{2} - 2} \times 4 t}{2 t}$

= $2 {e}^{2 {t}^{2} - 2}$