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