There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{(({a}^{2} + {b}^{2})(d - g) + ({c}^{2} + {d}^{2})(g - b) + ({f}^{2} + {g}^{2})(d - b))}{(2a(d - g) - 2b(c - f) + 2cg - 2fd)} - x\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{a^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{a^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{b^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{b^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{gc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{d^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bd^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{df^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bf^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{dg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - x\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{a^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{a^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{b^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{b^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{gc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{d^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bd^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{df^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bf^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{dg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - x\right)}{dx}\\=&(\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})a^{2}d + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})a^{2}g + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})b^{2}d + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})b^{2}g + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})gc^{2} + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bc^{2} + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})d^{2}g + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bd^{2} + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})df^{2} + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bf^{2} + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})dg^{2} + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bg^{2} + 0 - 1\\=& - 1\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!