There are 2 questions in this calculation: for each question, the 1 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/2]Find\ the\ first\ derivative\ of\ function\ \frac{{e}^{(2x)}(sin(x) + 2cos(x))}{5} + C\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{1}{5}{e}^{(2x)}sin(x) + \frac{2}{5}{e}^{(2x)}cos(x) + C\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{1}{5}{e}^{(2x)}sin(x) + \frac{2}{5}{e}^{(2x)}cos(x) + C\right)}{dx}\\=&\frac{1}{5}({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) + \frac{1}{5}{e}^{(2x)}cos(x) + \frac{2}{5}({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) + \frac{2}{5}{e}^{(2x)}*-sin(x) + 0\\=&{e}^{(2x)}cos(x)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[2/2]Find\ the\ first\ derivative\ of\ function\ \frac{{e}^{(2x)}(2sin(x) - cos(x))}{5} + C\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{2}{5}{e}^{(2x)}sin(x) - \frac{1}{5}{e}^{(2x)}cos(x) + C\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{2}{5}{e}^{(2x)}sin(x) - \frac{1}{5}{e}^{(2x)}cos(x) + C\right)}{dx}\\=&\frac{2}{5}({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) + \frac{2}{5}{e}^{(2x)}cos(x) - \frac{1}{5}({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - \frac{1}{5}{e}^{(2x)}*-sin(x) + 0\\=&{e}^{(2x)}sin(x)\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!