There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
Note that variables are case sensitive.\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ \frac{x}{(1 + sqrt(xx))}\ with\ respect\ to\ x:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{x}{(sqrt(x^{2}) + 1)}\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{x}{(sqrt(x^{2}) + 1)}\right)}{dx}\\=&(\frac{-(\frac{2x*\frac{1}{2}}{(x^{2})^{\frac{1}{2}}} + 0)}{(sqrt(x^{2}) + 1)^{2}})x + \frac{1}{(sqrt(x^{2}) + 1)}\\=&\frac{-x}{(sqrt(x^{2}) + 1)^{2}} + \frac{1}{(sqrt(x^{2}) + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-x}{(sqrt(x^{2}) + 1)^{2}} + \frac{1}{(sqrt(x^{2}) + 1)}\right)}{dx}\\=&-(\frac{-2(\frac{2x*\frac{1}{2}}{(x^{2})^{\frac{1}{2}}} + 0)}{(sqrt(x^{2}) + 1)^{3}})x - \frac{1}{(sqrt(x^{2}) + 1)^{2}} + (\frac{-(\frac{2x*\frac{1}{2}}{(x^{2})^{\frac{1}{2}}} + 0)}{(sqrt(x^{2}) + 1)^{2}})\\=&\frac{2x}{(sqrt(x^{2}) + 1)^{3}} - \frac{2}{(sqrt(x^{2}) + 1)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{2x}{(sqrt(x^{2}) + 1)^{3}} - \frac{2}{(sqrt(x^{2}) + 1)^{2}}\right)}{dx}\\=&2(\frac{-3(\frac{2x*\frac{1}{2}}{(x^{2})^{\frac{1}{2}}} + 0)}{(sqrt(x^{2}) + 1)^{4}})x + \frac{2}{(sqrt(x^{2}) + 1)^{3}} - 2(\frac{-2(\frac{2x*\frac{1}{2}}{(x^{2})^{\frac{1}{2}}} + 0)}{(sqrt(x^{2}) + 1)^{3}})\\=&\frac{-6x}{(sqrt(x^{2}) + 1)^{4}} + \frac{6}{(sqrt(x^{2}) + 1)^{3}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-6x}{(sqrt(x^{2}) + 1)^{4}} + \frac{6}{(sqrt(x^{2}) + 1)^{3}}\right)}{dx}\\=&-6(\frac{-4(\frac{2x*\frac{1}{2}}{(x^{2})^{\frac{1}{2}}} + 0)}{(sqrt(x^{2}) + 1)^{5}})x - \frac{6}{(sqrt(x^{2}) + 1)^{4}} + 6(\frac{-3(\frac{2x*\frac{1}{2}}{(x^{2})^{\frac{1}{2}}} + 0)}{(sqrt(x^{2}) + 1)^{4}})\\=&\frac{24x}{(sqrt(x^{2}) + 1)^{5}} - \frac{24}{(sqrt(x^{2}) + 1)^{4}}\\ \end{split}\end{equation} \]Your problem has not been solved here? Please take a look at the hot problems !