There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{(1 + 2sin(x)cos(x) - 4{sin(x)}^{2}cos(x) - 2sin(x))}{(1 + 2sin(x)cos(x) + 2{cos(x)}^{2})}\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{2sin(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{4sin^{2}(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{2sin(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} + \frac{1}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)}\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{2sin(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{4sin^{2}(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{2sin(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} + \frac{1}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)}\right)}{dx}\\=&2(\frac{-(2cos(x)cos(x) + 2sin(x)*-sin(x) + 2*-2cos(x)sin(x) + 0)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}})sin(x)cos(x) + \frac{2cos(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} + \frac{2sin(x)*-sin(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - 4(\frac{-(2cos(x)cos(x) + 2sin(x)*-sin(x) + 2*-2cos(x)sin(x) + 0)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}})sin^{2}(x)cos(x) - \frac{4*2sin(x)cos(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{4sin^{2}(x)*-sin(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - 2(\frac{-(2cos(x)cos(x) + 2sin(x)*-sin(x) + 2*-2cos(x)sin(x) + 0)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}})sin(x) - \frac{2cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} + (\frac{-(2cos(x)cos(x) + 2sin(x)*-sin(x) + 2*-2cos(x)sin(x) + 0)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}})\\=& - \frac{4sin(x)cos^{3}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} + \frac{8sin^{2}(x)cos^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} - \frac{16sin^{3}(x)cos^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} + \frac{2cos^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{8sin(x)cos^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} + \frac{8sin^{2}(x)cos^{3}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} - \frac{8sin^{4}(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} + \frac{4sin^{3}(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} + \frac{4sin(x)cos^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} - \frac{8sin^{2}(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} + \frac{4sin(x)cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} + \frac{4sin^{3}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{4sin^{3}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} - \frac{2cos(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} - \frac{2cos^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}} - \frac{2sin^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)} + \frac{2sin^{2}(x)}{(2sin(x)cos(x) + 2cos^{2}(x) + 1)^{2}}\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!