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

Dec 30, 2015

$f ' \left(t\right) = \frac{\sec \left(1 - t\right) \tan \left(1 - t\right)}{{e}^{t}}$

#### Explanation:

$x ' \left(t\right) = - {e}^{t}$
$y ' \left(t\right) = - \sec \left(1 - t\right) \tan \left(1 - t\right)$

The derivative of the parametric function is

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