There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{(1300 - x)(500 + x)}{(60000 + 40x)}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = - \frac{x^{2}}{(40x + 60000)} + \frac{800x}{(40x + 60000)} + \frac{650000}{(40x + 60000)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( - \frac{x^{2}}{(40x + 60000)} + \frac{800x}{(40x + 60000)} + \frac{650000}{(40x + 60000)}\right)}{dx}\\=& - (\frac{-(40 + 0)}{(40x + 60000)^{2}})x^{2} - \frac{2x}{(40x + 60000)} + 800(\frac{-(40 + 0)}{(40x + 60000)^{2}})x + \frac{800}{(40x + 60000)} + 650000(\frac{-(40 + 0)}{(40x + 60000)^{2}})\\=&\frac{40x^{2}}{(40x + 60000)^{2}} - \frac{2x}{(40x + 60000)} - \frac{32000x}{(40x + 60000)^{2}} - \frac{26000000}{(40x + 60000)^{2}} + \frac{800}{(40x + 60000)}\\ \end{split}\end{equation} \]

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