There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ lg(4 - x) - lg(4 + x)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = lg(-x + 4) - lg(x + 4)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( lg(-x + 4) - lg(x + 4)\right)}{dx}\\=&\frac{(-1 + 0)}{ln{10}(-x + 4)} - \frac{(1 + 0)}{ln{10}(x + 4)}\\=&\frac{-1}{(-x + 4)ln{10}} - \frac{1}{(x + 4)ln{10}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-1}{(-x + 4)ln{10}} - \frac{1}{(x + 4)ln{10}}\right)}{dx}\\=&\frac{-(\frac{-(-1 + 0)}{(-x + 4)^{2}})}{ln{10}} - \frac{-0}{(-x + 4)ln^{2}{10}} - \frac{(\frac{-(1 + 0)}{(x + 4)^{2}})}{ln{10}} - \frac{-0}{(x + 4)ln^{2}{10}}\\=&\frac{-1}{(-x + 4)^{2}ln{10}} + \frac{1}{(x + 4)^{2}ln{10}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-1}{(-x + 4)^{2}ln{10}} + \frac{1}{(x + 4)^{2}ln{10}}\right)}{dx}\\=&\frac{-(\frac{-2(-1 + 0)}{(-x + 4)^{3}})}{ln{10}} - \frac{-0}{(-x + 4)^{2}ln^{2}{10}} + \frac{(\frac{-2(1 + 0)}{(x + 4)^{3}})}{ln{10}} + \frac{-0}{(x + 4)^{2}ln^{2}{10}}\\=&\frac{-2}{(-x + 4)^{3}ln{10}} - \frac{2}{(x + 4)^{3}ln{10}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2}{(-x + 4)^{3}ln{10}} - \frac{2}{(x + 4)^{3}ln{10}}\right)}{dx}\\=&\frac{-2(\frac{-3(-1 + 0)}{(-x + 4)^{4}})}{ln{10}} - \frac{2*-0}{(-x + 4)^{3}ln^{2}{10}} - \frac{2(\frac{-3(1 + 0)}{(x + 4)^{4}})}{ln{10}} - \frac{2*-0}{(x + 4)^{3}ln^{2}{10}}\\=&\frac{-6}{(-x + 4)^{4}ln{10}} + \frac{6}{(x + 4)^{4}ln{10}}\\ \end{split}\end{equation} \]

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