There are 2 questions in this calculation: for each question, the 1 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/2]Find\ the\ first\ derivative\ of\ function\ \frac{5{e}^{(\frac{4x}{5})}(5sin(2x) + 2cos(2x))}{58} + C\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{25}{58}{e}^{(\frac{4}{5}x)}sin(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}cos(2x) + C\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{25}{58}{e}^{(\frac{4}{5}x)}sin(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}cos(2x) + C\right)}{dx}\\=&\frac{25}{58}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))sin(2x) + \frac{25}{58}{e}^{(\frac{4}{5}x)}cos(2x)*2 + \frac{5}{29}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))cos(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}*-sin(2x)*2 + 0\\=&{e}^{(\frac{4}{5}x)}cos(2x)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[2/2]Find\ the\ first\ derivative\ of\ function\ \frac{5{e}^{(\frac{4x}{5})}(2sin(2x) - 5cos(2x))}{58} + C\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{5}{29}{e}^{(\frac{4}{5}x)}sin(2x) - \frac{25}{58}{e}^{(\frac{4}{5}x)}cos(2x) + C\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{5}{29}{e}^{(\frac{4}{5}x)}sin(2x) - \frac{25}{58}{e}^{(\frac{4}{5}x)}cos(2x) + C\right)}{dx}\\=&\frac{5}{29}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))sin(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}cos(2x)*2 - \frac{25}{58}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))cos(2x) - \frac{25}{58}{e}^{(\frac{4}{5}x)}*-sin(2x)*2 + 0\\=&{e}^{(\frac{4}{5}x)}sin(2x)\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!