本次共计算 1 个题目：每一题对 t 求 1 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/1】求函数\frac{(p - c - D(t)(p + c + f))}{(D(t)(b - (p + c + f)))} 关于 t 的 1 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = - \frac{pDt}{(Dbt - pDt - cDt - Dft)} - \frac{cDt}{(Dbt - pDt - cDt - Dft)} + \frac{p}{(Dbt - pDt - cDt - Dft)} - \frac{c}{(Dbt - pDt - cDt - Dft)} - \frac{Dft}{(Dbt - pDt - cDt - Dft)}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( - \frac{pDt}{(Dbt - pDt - cDt - Dft)} - \frac{cDt}{(Dbt - pDt - cDt - Dft)} + \frac{p}{(Dbt - pDt - cDt - Dft)} - \frac{c}{(Dbt - pDt - cDt - Dft)} - \frac{Dft}{(Dbt - pDt - cDt - Dft)}\right)}{dt}\\=& - (\frac{-(Db - pD - cD - Df)}{(Dbt - pDt - cDt - Dft)^{2}})pDt - \frac{pD}{(Dbt - pDt - cDt - Dft)} - (\frac{-(Db - pD - cD - Df)}{(Dbt - pDt - cDt - Dft)^{2}})cDt - \frac{cD}{(Dbt - pDt - cDt - Dft)} + (\frac{-(Db - pD - cD - Df)}{(Dbt - pDt - cDt - Dft)^{2}})p + 0 - (\frac{-(Db - pD - cD - Df)}{(Dbt - pDt - cDt - Dft)^{2}})c + 0 - (\frac{-(Db - pD - cD - Df)}{(Dbt - pDt - cDt - Dft)^{2}})Dft - \frac{Df}{(Dbt - pDt - cDt - Dft)}\\=&\frac{pD^{2}bt}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{p^{2}D^{2}t}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{2pcD^{2}t}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{2pD^{2}ft}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{pDb}{(Dbt - pDt - cDt - Dft)^{2}} + \frac{cD^{2}bt}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{c^{2}D^{2}t}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{2cD^{2}ft}{(Dbt - pDt - cDt - Dft)^{2}} + \frac{cDb}{(Dbt - pDt - cDt - Dft)^{2}} + \frac{pDf}{(Dbt - pDt - cDt - Dft)^{2}} + \frac{p^{2}D}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{pD}{(Dbt - pDt - cDt - Dft)} - \frac{cDf}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{c^{2}D}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{cD}{(Dbt - pDt - cDt - Dft)} + \frac{D^{2}fbt}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{D^{2}f^{2}t}{(Dbt - pDt - cDt - Dft)^{2}} - \frac{Df}{(Dbt - pDt - cDt - Dft)}\\ \end{split}\end{equation} \]

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