Question $3$

$2$ pts

Given $y={\displaystyle \underset{0}{\overset{tanx}{}}}{e}^{-t^2}dt,$ then

$\frac{dy}{dx}=e^{-tan{x}^2}*se{c}^{2}x$

$\frac{dy}{dx}=e^{-x^2}*se{c}^{2}x$

$\frac{dy}{dx}=-e^{ta{n}^{2}x}*se{c}^{2}x$

$\frac{dy}{dx}=e^{-ta{n}^{2}x}*se{c}^{2}x$