There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
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\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ {10}^{{10}^{x}}{lg(10)}^{2022}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = {10}^{{10}^{x}}lg^{2022}(10)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( {10}^{{10}^{x}}lg^{2022}(10)\right)}{dx}\\=&({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))lg^{2022}(10) + \frac{{10}^{{10}^{x}}*2022lg^{2021}(10)*0}{ln{10}(10)}\\=&{10}^{x}{10}^{{10}^{x}}ln^{2}(10)lg^{2022}(10)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( {10}^{x}{10}^{{10}^{x}}ln^{2}(10)lg^{2022}(10)\right)}{dx}\\=&({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})){10}^{{10}^{x}}ln^{2}(10)lg^{2022}(10) + {10}^{x}({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))ln^{2}(10)lg^{2022}(10) + \frac{{10}^{x}{10}^{{10}^{x}}*2ln(10)*0lg^{2022}(10)}{(10)} + \frac{{10}^{x}{10}^{{10}^{x}}ln^{2}(10)*2022lg^{2021}(10)*0}{ln{10}(10)}\\=&{10}^{x}{10}^{{10}^{x}}ln^{3}(10)lg^{2022}(10) + {10}^{(2(x))}{10}^{{10}^{x}}ln^{4}(10)lg^{2022}(10)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( {10}^{x}{10}^{{10}^{x}}ln^{3}(10)lg^{2022}(10) + {10}^{(2(x))}{10}^{{10}^{x}}ln^{4}(10)lg^{2022}(10)\right)}{dx}\\=&({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})){10}^{{10}^{x}}ln^{3}(10)lg^{2022}(10) + {10}^{x}({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))ln^{3}(10)lg^{2022}(10) + \frac{{10}^{x}{10}^{{10}^{x}}*3ln^{2}(10)*0lg^{2022}(10)}{(10)} + \frac{{10}^{x}{10}^{{10}^{x}}ln^{3}(10)*2022lg^{2021}(10)*0}{ln{10}(10)} + ({10}^{(2(x))}((2(1))ln(10) + \frac{(2(x))(0)}{(10)})){10}^{{10}^{x}}ln^{4}(10)lg^{2022}(10) + {10}^{(2(x))}({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))ln^{4}(10)lg^{2022}(10) + \frac{{10}^{(2(x))}{10}^{{10}^{x}}*4ln^{3}(10)*0lg^{2022}(10)}{(10)} + \frac{{10}^{(2(x))}{10}^{{10}^{x}}ln^{4}(10)*2022lg^{2021}(10)*0}{ln{10}(10)}\\=&{10}^{x}{10}^{{10}^{x}}ln^{4}(10)lg^{2022}(10) + {10}^{(2(x))}{10}^{{10}^{x}}ln^{5}(10)lg^{2022}(10) + 2 * {10}^{(2x)}{10}^{{10}^{x}}ln^{5}(10)lg^{2022}(10) + {10}^{(3x)}{10}^{{10}^{x}}ln^{6}(10)lg^{2022}(10)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( {10}^{x}{10}^{{10}^{x}}ln^{4}(10)lg^{2022}(10) + {10}^{(2(x))}{10}^{{10}^{x}}ln^{5}(10)lg^{2022}(10) + 2 * {10}^{(2x)}{10}^{{10}^{x}}ln^{5}(10)lg^{2022}(10) + {10}^{(3x)}{10}^{{10}^{x}}ln^{6}(10)lg^{2022}(10)\right)}{dx}\\=&({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})){10}^{{10}^{x}}ln^{4}(10)lg^{2022}(10) + {10}^{x}({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))ln^{4}(10)lg^{2022}(10) + \frac{{10}^{x}{10}^{{10}^{x}}*4ln^{3}(10)*0lg^{2022}(10)}{(10)} + \frac{{10}^{x}{10}^{{10}^{x}}ln^{4}(10)*2022lg^{2021}(10)*0}{ln{10}(10)} + ({10}^{(2(x))}((2(1))ln(10) + \frac{(2(x))(0)}{(10)})){10}^{{10}^{x}}ln^{5}(10)lg^{2022}(10) + {10}^{(2(x))}({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))ln^{5}(10)lg^{2022}(10) + \frac{{10}^{(2(x))}{10}^{{10}^{x}}*5ln^{4}(10)*0lg^{2022}(10)}{(10)} + \frac{{10}^{(2(x))}{10}^{{10}^{x}}ln^{5}(10)*2022lg^{2021}(10)*0}{ln{10}(10)} + 2({10}^{(2x)}((2)ln(10) + \frac{(2x)(0)}{(10)})){10}^{{10}^{x}}ln^{5}(10)lg^{2022}(10) + 2 * {10}^{(2x)}({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))ln^{5}(10)lg^{2022}(10) + \frac{2 * {10}^{(2x)}{10}^{{10}^{x}}*5ln^{4}(10)*0lg^{2022}(10)}{(10)} + \frac{2 * {10}^{(2x)}{10}^{{10}^{x}}ln^{5}(10)*2022lg^{2021}(10)*0}{ln{10}(10)} + ({10}^{(3x)}((3)ln(10) + \frac{(3x)(0)}{(10)})){10}^{{10}^{x}}ln^{6}(10)lg^{2022}(10) + {10}^{(3x)}({10}^{{10}^{x}}((({10}^{x}((1)ln(10) + \frac{(x)(0)}{(10)})))ln(10) + \frac{({10}^{x})(0)}{(10)}))ln^{6}(10)lg^{2022}(10) + \frac{{10}^{(3x)}{10}^{{10}^{x}}*6ln^{5}(10)*0lg^{2022}(10)}{(10)} + \frac{{10}^{(3x)}{10}^{{10}^{x}}ln^{6}(10)*2022lg^{2021}(10)*0}{ln{10}(10)}\\=&{10}^{x}{10}^{{10}^{x}}ln^{5}(10)lg^{2022}(10) + {10}^{(2(x))}{10}^{{10}^{x}}ln^{6}(10)lg^{2022}(10) + 6 * {10}^{(2x)}{10}^{{10}^{x}}ln^{6}(10)lg^{2022}(10) + 6 * {10}^{(3x)}{10}^{{10}^{x}}ln^{7}(10)lg^{2022}(10) + {10}^{(4x)}{10}^{{10}^{x}}ln^{8}(10)lg^{2022}(10)\\ \end{split}\end{equation} \]

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