Answer
$\frac{{dy}}{{dx}} = {e^{ - {{\tan }^2}x}}{\sec ^2}x$
Work Step by Step
$$\eqalign{
& y = \int_0^{\tan x} {{e^{ - {t^2}}}dt} \cr
& {\text{Let }}u = \tan x,{\text{ we can use Part 1 of the Fundamental Theorem }} \cr
& {\text{of calculus with the chain rule as follows}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}} \cr
& {\text{Substituting }}y = \int_0^{\tan x} {{e^{ - {t^2}}}dt} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{du}}\left[ {\int_0^{\tan x} {{e^{ - {t^2}}}dt} } \right]\frac{{du}}{{dx}} \cr
& {\text{Substitute }}u{\text{ for }}\tan x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{du}}\left[ {\int_0^u {{e^{ - {t^2}}}dt} } \right]\frac{d}{{dx}}\left[ {\tan x} \right] \cr
& {\text{Use Part 1 of the Fundamental Theorem of calculus}} \cr
& \frac{{dy}}{{dx}} = {e^{ - {u^2}}}\left( {{{\sec }^2}x} \right) \cr
& {\text{Write in terms of }}x{\text{ substitute }}\tan x{\text{ for }}u \cr
& \frac{{dy}}{{dx}} = {e^{ - {{\tan }^2}x}}\left( {{{\sec }^2}x} \right) \cr
& \frac{{dy}}{{dx}} = {e^{ - {{\tan }^2}x}}{\sec ^2}x \cr} $$
