There are 4 questions in this calculation: for each question, the 4 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/4]Find\ the\ 4th\ derivative\ of\ function\ sin(x)\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( sin(x)\right)}{dx}\\=&cos(x)\\=&cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( cos(x)\right)}{dx}\\=&-sin(x)\\=&-sin(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( -sin(x)\right)}{dx}\\=&-cos(x)\\=&-cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -cos(x)\right)}{dx}\\=&--sin(x)\\=&sin(x)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[2/4]Find\ the\ 4th\ derivative\ of\ function\ sin(2x)\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( sin(2x)\right)}{dx}\\=&cos(2x)*2\\=&2cos(2x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 2cos(2x)\right)}{dx}\\=&2*-sin(2x)*2\\=&-4sin(2x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( -4sin(2x)\right)}{dx}\\=&-4cos(2x)*2\\=&-8cos(2x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -8cos(2x)\right)}{dx}\\=&-8*-sin(2x)*2\\=&16sin(2x)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[3/4]Find\ the\ 4th\ derivative\ of\ function\ sin(2x - 60)\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( sin(2x - 60)\right)}{dx}\\=&cos(2x - 60)(2 + 0)\\=&2cos(2x - 60)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 2cos(2x - 60)\right)}{dx}\\=&2*-sin(2x - 60)(2 + 0)\\=&-4sin(2x - 60)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( -4sin(2x - 60)\right)}{dx}\\=&-4cos(2x - 60)(2 + 0)\\=&-8cos(2x - 60)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -8cos(2x - 60)\right)}{dx}\\=&-8*-sin(2x - 60)(2 + 0)\\=&16sin(2x - 60)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[4/4]Find\ the\ 4th\ derivative\ of\ function\ 2sin(2x - 60)\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( 2sin(2x - 60)\right)}{dx}\\=&2cos(2x - 60)(2 + 0)\\=&4cos(2x - 60)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 4cos(2x - 60)\right)}{dx}\\=&4*-sin(2x - 60)(2 + 0)\\=&-8sin(2x - 60)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( -8sin(2x - 60)\right)}{dx}\\=&-8cos(2x - 60)(2 + 0)\\=&-16cos(2x - 60)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -16cos(2x - 60)\right)}{dx}\\=&-16*-sin(2x - 60)(2 + 0)\\=&32sin(2x - 60)\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!