Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 36

Determine the derivative of $f(t) = \csc ht (1 - \ln \csc ht)$. Simplify where possible.


$
\begin{equation}
\begin{aligned}

f'(t) =& \frac{d}{dt} \csc ht (1 - \ln \csc ht)
\\
\\
f'(t) =& (\csc ht ) \frac{d}{dt} (1 - \ln \csc ht) + (1 - \ln \csc ht) \frac{d}{dt} (\csc ht)
\\
\\
f'(t) =& (\cancel{\csc ht}) \left( \frac{-1}{\cancel{\csc ht}} \right) \cdot \frac{d}{dt} (\csc ht) + (1 - \ln \csc ht) (- \csc ht \cot ht)
\\
\\
f'(t) =& - (- \csc ht \cot ht) + (1 - \ln \csc ht) (- \csc ht \cot ht)
\\
\\
f'(t) =& \cancel{\csc ht \cot ht} - \cancel{\csc ht \cot ht} + \csc ht \cot ht \ln \csc ht
\\
\\
f'(t) =& \csc ht \cot ht \ln \csc ht

\end{aligned}
\end{equation}
$