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

Jan 10, 2018

$\left(3 {t}^{2} {e}^{1 - t} - {t}^{3} {e}^{1 - t}\right) i + 2 \tan t {\sec}^{2} t j$ Or,
$f ' \left(t\right) = \left(\left(3 {t}^{2} - {t}^{3}\right) {e}^{1 - t} , 2 \tan t {\sec}^{2} t\right)$

#### Explanation:

Given f(t) is ${t}^{3} {e}^{1 - t} i + {\tan}^{2} t j$

Its derivative f'(t) would be $\left(3 {t}^{2} {e}^{1 - t} - {t}^{3} {e}^{1 - t}\right) i + 2 \tan t {\sec}^{2} t j$. This can also be expressed as

$f ' \left(t\right) = \left(\left(3 {t}^{2} - {t}^{3}\right) {e}^{1 - t} , 2 \tan t {\sec}^{2} t\right)$