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

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