Answer
$$\frac{{dy}}{{dx}} = - 1$$
Work Step by Step
$$\eqalign{
& y = \int_{\tan x}^0 {\frac{{dt}}{{1 + {t^2}}}} \cr
& {\text{use the intergral property }}\cr
&\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx \cr
& y = - \int_0^{\tan x} {\frac{{dt}}{{1 + {t^2}}}} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = - \frac{d}{{dx}}\left( {\int_0^{\tan x} {\frac{{dt}}{{1 + {t^2}}}} } \right) \cr
& {\text{use the fundamental theorem of calculus part 1 }}\cr
&\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{1 + {{\left( {\tan x} \right)}^2}}}\frac{d}{{dx}}\left( {\tan x} \right) \cr
& {\text{compute the derivative and simplify}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{1 + {{\tan }^2}x}}\left( {{{\sec }^2}x} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\sec }^2}x}}\left( {{{\sec }^2}x} \right) \cr
& \frac{{dy}}{{dx}} = - 1 \cr} $$
