There are 4 questions in this calculation: for each question, the 4 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/4]Find\ the\ 4th\ derivative\ of\ function\ -cos(x)\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( -cos(x)\right)}{dx}\\=&--sin(x)\\=&sin(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( sin(x)\right)}{dx}\\=&cos(x)\\=&cos(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( cos(x)\right)}{dx}\\=&-sin(x)\\=&-sin(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -sin(x)\right)}{dx}\\=&-cos(x)\\=&-cos(x)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[2/4]Find\ the\ 4th\ derivative\ of\ function\ sin(x)\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( sin(x)\right)}{dx}\\=&cos(x)\\=&cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( cos(x)\right)}{dx}\\=&-sin(x)\\=&-sin(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( -sin(x)\right)}{dx}\\=&-cos(x)\\=&-cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -cos(x)\right)}{dx}\\=&--sin(x)\\=&sin(x)\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[3/4]Find\ the\ 4th\ derivative\ of\ function\ -ln(cos(x))\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( -ln(cos(x))\right)}{dx}\\=&\frac{--sin(x)}{(cos(x))}\\=&\frac{sin(x)}{cos(x)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{sin(x)}{cos(x)}\right)}{dx}\\=&\frac{cos(x)}{cos(x)} + \frac{sin(x)sin(x)}{cos^{2}(x)}\\=&\frac{sin^{2}(x)}{cos^{2}(x)} + 1\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{sin^{2}(x)}{cos^{2}(x)} + 1\right)}{dx}\\=&\frac{2sin(x)cos(x)}{cos^{2}(x)} + \frac{sin^{2}(x)*2sin(x)}{cos^{3}(x)} + 0\\=&\frac{2sin(x)}{cos(x)} + \frac{2sin^{3}(x)}{cos^{3}(x)}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{2sin(x)}{cos(x)} + \frac{2sin^{3}(x)}{cos^{3}(x)}\right)}{dx}\\=&\frac{2cos(x)}{cos(x)} + \frac{2sin(x)sin(x)}{cos^{2}(x)} + \frac{2*3sin^{2}(x)cos(x)}{cos^{3}(x)} + \frac{2sin^{3}(x)*3sin(x)}{cos^{4}(x)}\\=&\frac{8sin^{2}(x)}{cos^{2}(x)} + \frac{6sin^{4}(x)}{cos^{4}(x)} + 2\\ \end{split}\end{equation} \]\[ \begin{equation}\begin{split}[4/4]Find\ the\ 4th\ derivative\ of\ function\ ln(sin(x))\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( ln(sin(x))\right)}{dx}\\=&\frac{cos(x)}{(sin(x))}\\=&\frac{cos(x)}{sin(x)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{cos(x)}{sin(x)}\right)}{dx}\\=&\frac{-cos(x)cos(x)}{sin^{2}(x)} + \frac{-sin(x)}{sin(x)}\\=&\frac{-cos^{2}(x)}{sin^{2}(x)} - 1\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-cos^{2}(x)}{sin^{2}(x)} - 1\right)}{dx}\\=&\frac{--2cos(x)cos^{2}(x)}{sin^{3}(x)} - \frac{-2cos(x)sin(x)}{sin^{2}(x)} + 0\\=&\frac{2cos^{3}(x)}{sin^{3}(x)} + \frac{2cos(x)}{sin(x)}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{2cos^{3}(x)}{sin^{3}(x)} + \frac{2cos(x)}{sin(x)}\right)}{dx}\\=&\frac{2*-3cos(x)cos^{3}(x)}{sin^{4}(x)} + \frac{2*-3cos^{2}(x)sin(x)}{sin^{3}(x)} + \frac{2*-cos(x)cos(x)}{sin^{2}(x)} + \frac{2*-sin(x)}{sin(x)}\\=&\frac{-6cos^{4}(x)}{sin^{4}(x)} - \frac{8cos^{2}(x)}{sin^{2}(x)} - 2\\ \end{split}\end{equation} \]Your problem has not been solved here? Please go to the Hot Problems section!