## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {u^2}} }}$$
\eqalign{ & y = {\sin ^{ - 1}}\sqrt {1 - {u^2}},\,\,\,\,\,0 < u < 1 \cr & {\text{Find the derivative of }}y{\text{ with respect to }}u \cr & \frac{{dy}}{{dt}} = \frac{{d\left( {{{\sin }^{ - 1}}\sqrt {1 - {u^2}} } \right)}}{{dt}} \cr & {\text{we can use the formula }}\cr &\frac{{d\left( {{{\sin }^{ - 1}}z} \right)}}{{du}} = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{dz}}{{du}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr & {\text{here }}z = \sqrt {1 - {u^2}},\,\,{\text{then}} \cr & \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {1 - {{\left( {\sqrt {1 - {u^2}} } \right)}^2}} }}\frac{{d\left( {\sqrt {1 - {u^2}} } \right)}}{{du}} \cr & {\text{solving the derivative, we get:}} \cr & \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {1 - \left( {1 - {u^2}} \right)} }}\left( {\frac{{ - 2u}}{{2\sqrt {1 - {u^2}} }}} \right) \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {{u^2}} }}\left( {\frac{{ - u}}{{\sqrt {1 - {u^2}} }}} \right) \cr & \frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {u^2}} }} \cr}