## Thomas' Calculus 13th Edition

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