## Thomas' Calculus 13th Edition

$$\frac{{ds}}{{dt}} = 5\left( {6t - 1} \right){\csc ^5}\left( {1 - t + 3{t^2}} \right)\cot \left( {1 - t + 3{t^2}} \right)$$
\eqalign{ & s = {\csc ^5}\left( {1 - t + 3{t^2}} \right) \cr & {\text{differentiate with respect to }}t \cr & \frac{{ds}}{{dt}} = \frac{d}{{dt}}\left[ {{{\csc }^5}\left( {1 - t + 3{t^2}} \right)} \right] \cr & {\text{use the chain rule}} \cr & \frac{{ds}}{{dt}} = 5{\csc ^4}\left( {1 - t + 3{t^2}} \right)\frac{d}{{dt}}\left[ {{{\csc }^5}\left( {1 - t + 3{t^2}} \right)} \right] \cr & {\text{use }}\frac{d}{{dt}}\left[ {\csc u} \right] = - \csc u\cot u\frac{{du}}{{dx}} \cr & \frac{{ds}}{{dt}} = - 5{\csc ^4}\left( {1 - t + 3{t^2}} \right)\csc \left( {1 - t + 3{t^2}} \right)\cot \left( {1 - t + 3{t^2}} \right)\frac{d}{{dt}}\left[ {1 - t + 3{t^2}} \right] \cr & {\text{find the derivative and simplify}} \cr & \frac{{ds}}{{dt}} = - 5{\csc ^4}\left( {1 - t + 3{t^2}} \right)\csc \left( {1 - t + 3{t^2}} \right)\cot \left( {1 - t + 3{t^2}} \right)\left( { - 1 + 6t} \right) \cr & \frac{{ds}}{{dt}} = -5\left( {6t - 1} \right){\csc ^5}\left( {1 - t + 3{t^2}} \right)\cot \left( {1 - t + 3{t^2}} \right) \cr}