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

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