Derivative function：
Enter an original function (that is, the function to be derived), then set the variable to be derived and the order of the derivative, and click the "Next" button to obtain the derivative function of the corresponding order of the function.
Note that the input function supports mathematical functions and other constants.
Enter an original function (that is, the function to be derived), then set the variable to be derived and the order of the derivative, and click the "Next" button to obtain the derivative function of the corresponding order of the function.
Note that the input function supports mathematical functions and other constants.
Current location：Derivative function > Derivative function calculation history > Answer
There are 1 questions in this calculation: for each question, the 2 derivative of y is calculated.
Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ ({(\frac{sin(2x + y)}{4})}^{2}){\frac{1}{(2x + y)}}^{2} - ({(\frac{sin(2x - y)}{4})}^{2}){\frac{1}{(2x - y)}}^{2}\ with\ respect\ to\ y：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{1}{0}sin^{2}(2x + y)}{(2x + y)^{2}}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{1}{0}sin^{2}(2x + y)}{(2x + y)^{2}}\right)}{dy}\\=&\frac{1}{0}(\frac{-2(0 + 1)}{(2x + y)^{3}})sin^{2}(2x + y) + \frac{\frac{1}{0}*2sin(2x + y)cos(2x + y)(0 + 1)}{(2x + y)^{2}}\\=&\frac{2sin(2x + y)cos(2x + y)}{0(2x + y)^{2}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{3}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{2sin(2x + y)cos(2x + y)}{0(2x + y)^{2}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{3}}\right)}{dy}\\=&\frac{2(\frac{-2(0 + 1)}{(2x + y)^{3}})sin(2x + y)cos(2x + y)}{0} + \frac{2cos(2x + y)(0 + 1)cos(2x + y)}{0(2x + y)^{2}} + \frac{2sin(2x + y)*-sin(2x + y)(0 + 1)}{0(2x + y)^{2}} - \frac{2(\frac{-3(0 + 1)}{(2x + y)^{4}})sin^{2}(2x + y)}{0} - \frac{2*2sin(2x + y)cos(2x + y)(0 + 1)}{0(2x + y)^{3}}\\=&\frac{2cos^{2}(2x + y)}{0(2x + y)^{2}} - \frac{4sin(2x + y)cos(2x + y)}{0(2x + y)^{3}} + \frac{6sin^{2}(2x + y)}{0(2x + y)^{4}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{2}}\\ \end{split}\end{equation} \]
Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ ({(\frac{sin(2x + y)}{4})}^{2}){\frac{1}{(2x + y)}}^{2} - ({(\frac{sin(2x - y)}{4})}^{2}){\frac{1}{(2x - y)}}^{2}\ with\ respect\ to\ y：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{1}{0}sin^{2}(2x + y)}{(2x + y)^{2}}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{1}{0}sin^{2}(2x + y)}{(2x + y)^{2}}\right)}{dy}\\=&\frac{1}{0}(\frac{-2(0 + 1)}{(2x + y)^{3}})sin^{2}(2x + y) + \frac{\frac{1}{0}*2sin(2x + y)cos(2x + y)(0 + 1)}{(2x + y)^{2}}\\=&\frac{2sin(2x + y)cos(2x + y)}{0(2x + y)^{2}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{3}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{2sin(2x + y)cos(2x + y)}{0(2x + y)^{2}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{3}}\right)}{dy}\\=&\frac{2(\frac{-2(0 + 1)}{(2x + y)^{3}})sin(2x + y)cos(2x + y)}{0} + \frac{2cos(2x + y)(0 + 1)cos(2x + y)}{0(2x + y)^{2}} + \frac{2sin(2x + y)*-sin(2x + y)(0 + 1)}{0(2x + y)^{2}} - \frac{2(\frac{-3(0 + 1)}{(2x + y)^{4}})sin^{2}(2x + y)}{0} - \frac{2*2sin(2x + y)cos(2x + y)(0 + 1)}{0(2x + y)^{3}}\\=&\frac{2cos^{2}(2x + y)}{0(2x + y)^{2}} - \frac{4sin(2x + y)cos(2x + y)}{0(2x + y)^{3}} + \frac{6sin^{2}(2x + y)}{0(2x + y)^{4}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{2}}\\ \end{split}\end{equation} \]