## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dt}} = - \frac{{2 + 2{t^2}}}{{1 + 4{t^2}}} + 2t{\cot ^{ - 1}}2t$$
\eqalign{ & y = \left( {1 + {t^2}} \right){\cot ^{ - 1}}2t \cr & {\text{find the derivative of }}y{\text{ with respect to }}t \cr & \frac{{dy}}{{dt}} = \frac{{d\left( {\left( {1 + {t^2}} \right){{\cot }^{ - 1}}2t} \right)}}{{dt}} \cr & {\text{use the product rule}} \cr & \frac{{dy}}{{dt}} = \left( {1 + {t^2}} \right)\frac{{d\left( {{{\cot }^{ - 1}}2t} \right)}}{{dt}} + {\cot ^{ - 1}}2t\frac{{d\left( {1 + {t^2}} \right)}}{{dt}} \cr & {\text{use the formula }}\cr &\frac{{d\left( {{{\cot }^{ - 1}}u} \right)}}{{dt}} = - \frac{1}{{1 + {u^2}}}\frac{{du}}{{dt}} \cr & \frac{{dy}}{{dt}} = \left( {1 + {t^2}} \right)\left( { - \frac{1}{{1 + {{\left( {2t} \right)}^2}}}} \right)\frac{d}{{dt}}\left( {2t} \right) + {\cot ^{ - 1}}2t\left( {2t} \right) \cr & \frac{{dy}}{{dt}} = \left( {1 + {t^2}} \right)\left( { - \frac{1}{{1 + 4{t^2}}}} \right)\left( 2 \right) + {\cot ^{ - 1}}2t\left( {2t} \right) \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dt}} = - \frac{{2 + 2{t^2}}}{{1 + 4{t^2}}} + 2t{\cot ^{ - 1}}2t \cr}