There are 1 questions in this calculation: for each question, the 4 derivative of ln is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ ({({ln(x)}^{x}(e^{x}))}^{2})\ with\ respect\ to\ ln：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = {ln(x)}^{(2x)}e^{{x}*{2}}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( {ln(x)}^{(2x)}e^{{x}*{2}}\right)}{dln}\\=&({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}} + {ln(x)}^{(2x)}*2e^{x}e^{x}*0\\=&\frac{2x{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{2x{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)}\right)}{dln}\\=&\frac{2x*-{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{2}} + \frac{2x({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}}}{ln(x)} + \frac{2x{ln(x)}^{(2x)}*2e^{x}e^{x}*0}{ln(x)}\\=&\frac{-2x{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{2}} + \frac{4x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2x{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{2}} + \frac{4x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)^{2}}\right)}{dln}\\=&\frac{-2x*-2{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{3}} - \frac{2x({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}}}{ln^{2}} - \frac{2x{ln(x)}^{(2x)}*2e^{x}e^{x}*0}{ln^{2}} + \frac{4x^{2}*-2{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{3}} + \frac{4x^{2}({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}}}{ln(x)^{2}} + \frac{4x^{2}{ln(x)}^{(2x)}*2e^{x}e^{x}*0}{ln(x)^{2}}\\=&\frac{4x{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{3}} - \frac{4x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)ln^{2}} - \frac{8x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{3}} + \frac{8x^{3}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)^{3}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{4x{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{3}} - \frac{4x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)ln^{2}} - \frac{8x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{3}} + \frac{8x^{3}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)^{3}}\right)}{dln}\\=&\frac{4x*-3{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{4}} + \frac{4x({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}}}{ln^{3}} + \frac{4x{ln(x)}^{(2x)}*2e^{x}e^{x}*0}{ln^{3}} - \frac{4x^{2}*-{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{2}ln^{2}} - \frac{4x^{2}*-2{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)ln^{3}} - \frac{4x^{2}({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}}}{ln(x)ln^{2}} - \frac{4x^{2}{ln(x)}^{(2x)}*2e^{x}e^{x}*0}{ln(x)ln^{2}} - \frac{8x^{2}*-3{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{4}} - \frac{8x^{2}({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}}}{ln^{3}} - \frac{8x^{2}{ln(x)}^{(2x)}*2e^{x}e^{x}*0}{ln^{3}} + \frac{8x^{3}*-3{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{4}} + \frac{8x^{3}({ln(x)}^{(2x)}((0)ln(ln(x)) + \frac{(2x)(1)}{(ln(x))}))e^{{x}*{2}}}{ln(x)^{3}} + \frac{8x^{3}{ln(x)}^{(2x)}*2e^{x}e^{x}*0}{ln(x)^{3}}\\=&\frac{-12x{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{4}} + \frac{16x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)ln^{3}} + \frac{28x^{2}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{4}} - \frac{8x^{3}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)^{2}ln^{2}} - \frac{16x^{3}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)ln^{3}} - \frac{24x^{3}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln^{4}} + \frac{16x^{4}{ln(x)}^{(2x)}e^{{x}*{2}}}{ln(x)^{4}}\\ \end{split}\end{equation} \]

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