There are 1 questions in this calculation: for each question, the 1 derivative of k is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{{(1 - p)}^{k}}{k} + (1 - {(1 - p)}^{k})(1 + \frac{1}{k} + \frac{1}{k} - {(1 - p)}^{k})\ with\ respect\ to\ k:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{-(-p + 1)^{k}}{k} + \frac{2}{k} + (-p + 1)^{(2(k))} - 2(-p + 1)^{k} + 1\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{-(-p + 1)^{k}}{k} + \frac{2}{k} + (-p + 1)^{(2(k))} - 2(-p + 1)^{k} + 1\right)}{dk}\\=&\frac{--(-p + 1)^{k}}{k^{2}} - \frac{((-p + 1)^{k}((1)ln(-p + 1) + \frac{(k)(0 + 0)}{(-p + 1)}))}{k} + \frac{2*-1}{k^{2}} + ((-p + 1)^{(2(k))}((2(1))ln(-p + 1) + \frac{(2(k))(0 + 0)}{(-p + 1)})) - 2((-p + 1)^{k}((1)ln(-p + 1) + \frac{(k)(0 + 0)}{(-p + 1)})) + 0\\=& - \frac{(-p + 1)^{k}ln(-p + 1)}{k} + \frac{(-p + 1)^{k}}{k^{2}} - \frac{2}{k^{2}} + 2(-p + 1)^{(2k)}ln(-p + 1) - 2(-p + 1)^{k}ln(-p + 1)\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!