There are 1 questions in this calculation: for each question, the 1 derivative of z is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ sqrt(\frac{({(40 - z)}^{2} + 900)}{({(40 + z)}^{2} + 900)})\ with\ respect\ to\ z：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = sqrt(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( sqrt(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})\right)}{dz}\\=&\frac{((\frac{-(2z + 80 + 0)}{(z^{2} + 80z + 2500)^{2}})z^{2} + \frac{2z}{(z^{2} + 80z + 2500)} - 80(\frac{-(2z + 80 + 0)}{(z^{2} + 80z + 2500)^{2}})z - \frac{80}{(z^{2} + 80z + 2500)} + 2500(\frac{-(2z + 80 + 0)}{(z^{2} + 80z + 2500)^{2}}))*\frac{1}{2}}{(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})^{\frac{1}{2}}}\\=&\frac{-z^{3}}{(z^{2} + 80z + 2500)^{2}(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})^{\frac{1}{2}}} + \frac{40z^{2}}{(z^{2} + 80z + 2500)^{2}(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})^{\frac{1}{2}}} + \frac{z}{(z^{2} + 80z + 2500)(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})^{\frac{1}{2}}} + \frac{700z}{(z^{2} + 80z + 2500)^{2}(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})^{\frac{1}{2}}} - \frac{40}{(z^{2} + 80z + 2500)(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})^{\frac{1}{2}}} - \frac{100000}{(z^{2} + 80z + 2500)^{2}(\frac{z^{2}}{(z^{2} + 80z + 2500)} - \frac{80z}{(z^{2} + 80z + 2500)} + \frac{2500}{(z^{2} + 80z + 2500)})^{\frac{1}{2}}}\\ \end{split}\end{equation} \]

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