There are 2 questions in this calculation: for each question, the 4 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/2]Find\ the\ 4th\ derivative\ of\ function\ {sin(x)}^{2} - {cos(x)}^{2}\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = sin^{2}(x) - cos^{2}(x)\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( sin^{2}(x) - cos^{2}(x)\right)}{dx}\\=&2sin(x)cos(x) - -2cos(x)sin(x)\\=&4sin(x)cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 4sin(x)cos(x)\right)}{dx}\\=&4cos(x)cos(x) + 4sin(x)*-sin(x)\\=&4cos^{2}(x) - 4sin^{2}(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( 4cos^{2}(x) - 4sin^{2}(x)\right)}{dx}\\=&4*-2cos(x)sin(x) - 4*2sin(x)cos(x)\\=&-16sin(x)cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -16sin(x)cos(x)\right)}{dx}\\=&-16cos(x)cos(x) - 16sin(x)*-sin(x)\\=&-16cos^{2}(x) + 16sin^{2}(x)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[2/2]Find\ the\ 4th\ derivative\ of\ function\ {sin(x)}^{4} - {cos(x)}^{4}\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = sin^{4}(x) - cos^{4}(x)\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( sin^{4}(x) - cos^{4}(x)\right)}{dx}\\=&4sin^{3}(x)cos(x) - -4cos^{3}(x)sin(x)\\=&4sin^{3}(x)cos(x) + 4sin(x)cos^{3}(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 4sin^{3}(x)cos(x) + 4sin(x)cos^{3}(x)\right)}{dx}\\=&4*3sin^{2}(x)cos(x)cos(x) + 4sin^{3}(x)*-sin(x) + 4cos(x)cos^{3}(x) + 4sin(x)*-3cos^{2}(x)sin(x)\\=&4cos^{4}(x) - 4sin^{4}(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( 4cos^{4}(x) - 4sin^{4}(x)\right)}{dx}\\=&4*-4cos^{3}(x)sin(x) - 4*4sin^{3}(x)cos(x)\\=& - 16sin(x)cos^{3}(x) - 16sin^{3}(x)cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( - 16sin(x)cos^{3}(x) - 16sin^{3}(x)cos(x)\right)}{dx}\\=& - 16cos(x)cos^{3}(x) - 16sin(x)*-3cos^{2}(x)sin(x) - 16*3sin^{2}(x)cos(x)cos(x) - 16sin^{3}(x)*-sin(x)\\=& - 16cos^{4}(x) + 16sin^{4}(x)\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!