Answer
\[\frac{{dy}}{{dt}} = - \,\left( {2t + 1} \right)\csc \,\left( {{t^2} + t} \right)\cot \,\left( {{t^2} + t} \right)\]
Work Step by Step
\[\begin{gathered}
y = \csc \,\left( {{t^2} + t} \right) \hfill \\
\hfill \\
Use\,\,the\,\,version\,\,2\,\,of\,\,the\,\,chain\,\,rule\, \hfill \\
\hfill \\
\,\,\frac{d}{{dx}}\,\,\left[ {f\,\left( {g\,\left( x \right)} \right)} \right] = {f^,}\,\left( {g\,\left( x \right)} \right) \cdot {g^,}\,\left( x \right) \hfill \\
\hfill \\
set\,\,\,g\,\left( t \right) = {t^2} + t \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\,\,\,\,\,\,\,\,\,\,\,{g^,}\,\left( t \right) = 2t + 1 \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = - \csc \,\left( {{t^2} + t} \right)\cot \,\left( {{t^2} + t} \right)\,\left( {2t + 1} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dt}} = - \,\left( {2t + 1} \right)\csc \,\left( {{t^2} + t} \right)\cot \,\left( {{t^2} + t} \right) \hfill \\
\hfill \\
\end{gathered} \]
