There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ xsin({x}^{2}) - 2(1 - cos(x))sin(x)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = xsin(x^{2}) + 2sin(x)cos(x) - 2sin(x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( xsin(x^{2}) + 2sin(x)cos(x) - 2sin(x)\right)}{dx}\\=&sin(x^{2}) + xcos(x^{2})*2x + 2cos(x)cos(x) + 2sin(x)*-sin(x) - 2cos(x)\\=&sin(x^{2}) + 2x^{2}cos(x^{2}) + 2cos^{2}(x) - 2sin^{2}(x) - 2cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( sin(x^{2}) + 2x^{2}cos(x^{2}) + 2cos^{2}(x) - 2sin^{2}(x) - 2cos(x)\right)}{dx}\\=&cos(x^{2})*2x + 2*2xcos(x^{2}) + 2x^{2}*-sin(x^{2})*2x + 2*-2cos(x)sin(x) - 2*2sin(x)cos(x) - 2*-sin(x)\\=&6xcos(x^{2}) - 4x^{3}sin(x^{2}) - 8sin(x)cos(x) + 2sin(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( 6xcos(x^{2}) - 4x^{3}sin(x^{2}) - 8sin(x)cos(x) + 2sin(x)\right)}{dx}\\=&6cos(x^{2}) + 6x*-sin(x^{2})*2x - 4*3x^{2}sin(x^{2}) - 4x^{3}cos(x^{2})*2x - 8cos(x)cos(x) - 8sin(x)*-sin(x) + 2cos(x)\\=&6cos(x^{2}) - 24x^{2}sin(x^{2}) - 8x^{4}cos(x^{2}) - 8cos^{2}(x) + 8sin^{2}(x) + 2cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( 6cos(x^{2}) - 24x^{2}sin(x^{2}) - 8x^{4}cos(x^{2}) - 8cos^{2}(x) + 8sin^{2}(x) + 2cos(x)\right)}{dx}\\=&6*-sin(x^{2})*2x - 24*2xsin(x^{2}) - 24x^{2}cos(x^{2})*2x - 8*4x^{3}cos(x^{2}) - 8x^{4}*-sin(x^{2})*2x - 8*-2cos(x)sin(x) + 8*2sin(x)cos(x) + 2*-sin(x)\\=&-60xsin(x^{2}) - 80x^{3}cos(x^{2}) + 16x^{5}sin(x^{2}) + 32sin(x)cos(x) - 2sin(x)\\ \end{split}\end{equation} \]

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