There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{(\frac{-k{h}^{3}}{3} + (71k - \frac{1}{2})x + 142b)}{(kx + b)}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{-1}{3}kh^{3}}{(kx + b)} + \frac{71kx}{(kx + b)} - \frac{\frac{1}{2}x}{(kx + b)} + \frac{142b}{(kx + b)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{-1}{3}kh^{3}}{(kx + b)} + \frac{71kx}{(kx + b)} - \frac{\frac{1}{2}x}{(kx + b)} + \frac{142b}{(kx + b)}\right)}{dx}\\=&\frac{-1}{3}(\frac{-(k + 0)}{(kx + b)^{2}})kh^{3} + 0 + 71(\frac{-(k + 0)}{(kx + b)^{2}})kx + \frac{71k}{(kx + b)} - \frac{1}{2}(\frac{-(k + 0)}{(kx + b)^{2}})x - \frac{\frac{1}{2}}{(kx + b)} + 142(\frac{-(k + 0)}{(kx + b)^{2}})b + 0\\=&\frac{k^{2}h^{3}}{3(kx + b)^{2}} - \frac{71k^{2}x}{(kx + b)^{2}} + \frac{kx}{2(kx + b)^{2}} - \frac{142kb}{(kx + b)^{2}} + \frac{71k}{(kx + b)} - \frac{1}{2(kx + b)}\\ \end{split}\end{equation} \]

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