There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{(xA + (1 - x)B - D)}{sqrt(({x}^{2})({A}^{2}) + ({(1 - x)}^{2}){B}^{2} + 2Cx(1 - x)AB)}\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{Ax}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} + \frac{B}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} - \frac{Bx}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} - \frac{D}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})}\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{Ax}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} + \frac{B}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} - \frac{Bx}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} - \frac{D}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})}\right)}{dx}\\=&\frac{A}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} + \frac{Ax*-(A^{2}*2x + B^{2}*2x - 2B^{2} + 0 + 2ABC - 2ABC*2x)*\frac{1}{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{1}{2}}} + \frac{B*-(A^{2}*2x + B^{2}*2x - 2B^{2} + 0 + 2ABC - 2ABC*2x)*\frac{1}{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{1}{2}}} - \frac{B}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} - \frac{Bx*-(A^{2}*2x + B^{2}*2x - 2B^{2} + 0 + 2ABC - 2ABC*2x)*\frac{1}{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{1}{2}}} - \frac{D*-(A^{2}*2x + B^{2}*2x - 2B^{2} + 0 + 2ABC - 2ABC*2x)*\frac{1}{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{1}{2}}}\\=&\frac{A}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} - \frac{A^{3}x^{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{AB^{2}x^{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{AB^{2}x}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{A^{2}BCx}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{2A^{2}BCx^{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{A^{2}Bx^{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{B^{3}x^{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{2B^{3}x}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{3AB^{2}Cx}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{2AB^{2}Cx^{2}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{B}{sqrt(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})} - \frac{A^{2}Bx}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{B^{2}Dx}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{AB^{2}C}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{A^{2}Dx}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{B^{2}D}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{B^{3}}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} - \frac{2ABDCx}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}} + \frac{ABDC}{(A^{2}x^{2} + B^{2}x^{2} - 2B^{2}x + B^{2} + 2ABCx - 2ABCx^{2})^{\frac{3}{2}}}\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!