There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{ln(sin(x))}{(2cos(x)sin(x))} - \frac{1}{(2ln(sin(x)))}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{1}{2}ln(sin(x))}{sin(x)cos(x)} - \frac{\frac{1}{2}}{ln(sin(x))}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{1}{2}ln(sin(x))}{sin(x)cos(x)} - \frac{\frac{1}{2}}{ln(sin(x))}\right)}{dx}\\=&\frac{\frac{1}{2}cos(x)}{(sin(x))sin(x)cos(x)} + \frac{\frac{1}{2}ln(sin(x))*-cos(x)}{sin^{2}(x)cos(x)} + \frac{\frac{1}{2}ln(sin(x))sin(x)}{sin(x)cos^{2}(x)} - \frac{\frac{1}{2}*-cos(x)}{ln^{2}(sin(x))(sin(x))}\\=&\frac{1}{2sin^{2}(x)} + \frac{cos(x)}{2ln^{2}(sin(x))sin(x)} + \frac{ln(sin(x))}{2cos^{2}(x)} - \frac{ln(sin(x))}{2sin^{2}(x)}\\ \end{split}\end{equation} \]

Your problem has not been solved here? Please take a look at the  hot problems ！