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

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