# What is the derivative of f(t) = (t^3-e^(1-t) , tan^2t ) ?

Mar 31, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {e}^{2} + {e}^{1 - t}}{2 \tan t \cdot {\sec}^{2} t}$.

#### Explanation:

$f \left(t\right) = \left({t}^{3} - {e}^{1 - t} , {\tan}^{2} t\right)$ is a parametric function in which both $x$ and $y$ are function of $t$.

As such $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$.

Here $\frac{\mathrm{dy}}{\mathrm{dt}} = 3 {e}^{2} - \left(- 1\right) \cdot {e}^{1 - t} = 3 {e}^{2} + {e}^{1 - t}$ and

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 \tan t \cdot {\sec}^{2} t$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {e}^{2} + {e}^{1 - t}}{2 \tan t \cdot {\sec}^{2} t}$.