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