There are 1 questions in this calculation: for each question, the 2 derivative of z is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ {(aa + zz)}^{\frac{-3}{2}} - {(aa + (z - b)(z - b))}^{\frac{-3}{2}}\ with\ respect\ to\ z：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{1}{(a^{2} + z^{2})^{\frac{3}{2}}} - \frac{1}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{3}{2}}}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{1}{(a^{2} + z^{2})^{\frac{3}{2}}} - \frac{1}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{3}{2}}}\right)}{dz}\\=&(\frac{\frac{-3}{2}(0 + 2z)}{(a^{2} + z^{2})^{\frac{5}{2}}}) - (\frac{\frac{-3}{2}(0 + 2z - 2b + 0)}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{5}{2}}})\\=&\frac{-3z}{(a^{2} + z^{2})^{\frac{5}{2}}} + \frac{3z}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{5}{2}}} - \frac{3b}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{5}{2}}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-3z}{(a^{2} + z^{2})^{\frac{5}{2}}} + \frac{3z}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{5}{2}}} - \frac{3b}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{5}{2}}}\right)}{dz}\\=&-3(\frac{\frac{-5}{2}(0 + 2z)}{(a^{2} + z^{2})^{\frac{7}{2}}})z - \frac{3}{(a^{2} + z^{2})^{\frac{5}{2}}} + 3(\frac{\frac{-5}{2}(0 + 2z - 2b + 0)}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{7}{2}}})z + \frac{3}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{5}{2}}} - 3(\frac{\frac{-5}{2}(0 + 2z - 2b + 0)}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{7}{2}}})b + 0\\=&\frac{15z^{2}}{(a^{2} + z^{2})^{\frac{7}{2}}} - \frac{15z^{2}}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{7}{2}}} + \frac{30bz}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{7}{2}}} - \frac{15b^{2}}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{7}{2}}} + \frac{3}{(a^{2} + z^{2} - 2bz + b^{2})^{\frac{5}{2}}} - \frac{3}{(a^{2} + z^{2})^{\frac{5}{2}}}\\ \end{split}\end{equation} \]

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