There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ {2}^{(x + y)}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( {2}^{(x + y)}\right)}{dx}\\=&({2}^{(x + y)}((1 + 0)ln(2) + \frac{(x + y)(0)}{(2)}))\\=&{2}^{(x + y)}ln(2)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( {2}^{(x + y)}ln(2)\right)}{dx}\\=&({2}^{(x + y)}((1 + 0)ln(2) + \frac{(x + y)(0)}{(2)}))ln(2) + \frac{{2}^{(x + y)}*0}{(2)}\\=&{2}^{(x + y)}ln^{2}(2)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( {2}^{(x + y)}ln^{2}(2)\right)}{dx}\\=&({2}^{(x + y)}((1 + 0)ln(2) + \frac{(x + y)(0)}{(2)}))ln^{2}(2) + \frac{{2}^{(x + y)}*2ln(2)*0}{(2)}\\=&{2}^{(x + y)}ln^{3}(2)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( {2}^{(x + y)}ln^{3}(2)\right)}{dx}\\=&({2}^{(x + y)}((1 + 0)ln(2) + \frac{(x + y)(0)}{(2)}))ln^{3}(2) + \frac{{2}^{(x + y)}*3ln^{2}(2)*0}{(2)}\\=&{2}^{(x + y)}ln^{4}(2)\\ \end{split}\end{equation} \]

Your problem has not been solved here? Please take a look at the  hot problems ！