## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = - \frac{2}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 2} }}$$
\eqalign{ & y = {\csc ^{ - 1}}\left( {{x^2} + 1} \right),\,\,\,\,x > 0 \cr & {\text{find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{{d\left( {{{\csc }^{ - 1}}\left( {{x^2} + 1} \right)} \right)}}{{dx}} \cr & {\text{we can use the formula }}\cr &\frac{{d\left( {{{\csc }^{ - 1}}u} \right)}}{{dx}} = - \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr & {\text{here }}u = {x^2} + 1;{\text{then}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{\left| {{x^2} + 1} \right|\sqrt {{{\left( {{x^2} + 1} \right)}^2} - 1} }}\frac{{d\left( {{x^2} + 1} \right)}}{{dx}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{\left( {{x^2} + 1} \right)\sqrt {{x^4} + 2{x^2} + 1 - 1} }}\left( {2x} \right) \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dx}} = - \frac{{2x}}{{\left( {{x^2} + 1} \right)\sqrt {{x^4} + 2{x^2}} }} \cr & \frac{{dy}}{{dx}} = - \frac{{2x}}{{\left( {{x^2} + 1} \right)\left( x \right)\sqrt {{x^2} + 2} }} \cr & \frac{{dy}}{{dx}} = - \frac{2}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 2} }} \cr}