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 2 questions in this calculation: for each question, the 4 derivative of b is calculated.
Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/2]Find\ the\ 4th\ derivative\ of\ function\ \frac{(-b + sqrt({b}^{2} - 4ac))}{(2a)}\ with\ respect\ to\ b：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{-1}{2}b}{a} + \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{-1}{2}b}{a} + \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\right)}{db}\\=&\frac{\frac{-1}{2}}{a} + \frac{\frac{1}{2}(2b + 0)*\frac{1}{2}}{a(b^{2} - 4ac)^{\frac{1}{2}}}\\=&\frac{-1}{2a} + \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-1}{2a} + \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&0 + \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})b}{2a} + \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\=&\frac{-b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} + \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} + \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&\frac{-(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b^{2}}{2a} - \frac{2b}{2(b^{2} - 4ac)^{\frac{3}{2}}a} + \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})}{2a} + 0\\=&\frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\right)}{db}\\=&\frac{3(\frac{\frac{-5}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{7}{2}}})b^{3}}{2a} + \frac{3*3b^{2}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b}{2a} - \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\=&\frac{-15b^{4}}{2(b^{2} - 4ac)^{\frac{7}{2}}a} + \frac{9b^{2}}{(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\ \end{split}\end{equation} \]
\[ \begin{equation}\begin{split}[2/2]Find\ the\ 4th\ derivative\ of\ function\ \frac{(-b - sqrt({b}^{2} - 4ac))}{(2a)}\ with\ respect\ to\ b：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{-1}{2}b}{a} - \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{-1}{2}b}{a} - \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\right)}{db}\\=&\frac{\frac{-1}{2}}{a} - \frac{\frac{1}{2}(2b + 0)*\frac{1}{2}}{a(b^{2} - 4ac)^{\frac{1}{2}}}\\=&\frac{-1}{2a} - \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-1}{2a} - \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&0 - \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})b}{2a} - \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\=&\frac{b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} - \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} - \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&\frac{(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b^{2}}{2a} + \frac{2b}{2(b^{2} - 4ac)^{\frac{3}{2}}a} - \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})}{2a} + 0\\=& - \frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( - \frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\right)}{db}\\=& - \frac{3(\frac{\frac{-5}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{7}{2}}})b^{3}}{2a} - \frac{3*3b^{2}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b}{2a} + \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\=&\frac{15b^{4}}{2(b^{2} - 4ac)^{\frac{7}{2}}a} - \frac{9b^{2}}{(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\ \end{split}\end{equation} \]
Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/2]Find\ the\ 4th\ derivative\ of\ function\ \frac{(-b + sqrt({b}^{2} - 4ac))}{(2a)}\ with\ respect\ to\ b：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{-1}{2}b}{a} + \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{-1}{2}b}{a} + \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\right)}{db}\\=&\frac{\frac{-1}{2}}{a} + \frac{\frac{1}{2}(2b + 0)*\frac{1}{2}}{a(b^{2} - 4ac)^{\frac{1}{2}}}\\=&\frac{-1}{2a} + \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-1}{2a} + \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&0 + \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})b}{2a} + \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\=&\frac{-b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} + \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} + \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&\frac{-(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b^{2}}{2a} - \frac{2b}{2(b^{2} - 4ac)^{\frac{3}{2}}a} + \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})}{2a} + 0\\=&\frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\right)}{db}\\=&\frac{3(\frac{\frac{-5}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{7}{2}}})b^{3}}{2a} + \frac{3*3b^{2}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b}{2a} - \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\=&\frac{-15b^{4}}{2(b^{2} - 4ac)^{\frac{7}{2}}a} + \frac{9b^{2}}{(b^{2} - 4ac)^{\frac{5}{2}}a} - \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\ \end{split}\end{equation} \]
\[ \begin{equation}\begin{split}[2/2]Find\ the\ 4th\ derivative\ of\ function\ \frac{(-b - sqrt({b}^{2} - 4ac))}{(2a)}\ with\ respect\ to\ b：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{-1}{2}b}{a} - \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{-1}{2}b}{a} - \frac{\frac{1}{2}sqrt(b^{2} - 4ac)}{a}\right)}{db}\\=&\frac{\frac{-1}{2}}{a} - \frac{\frac{1}{2}(2b + 0)*\frac{1}{2}}{a(b^{2} - 4ac)^{\frac{1}{2}}}\\=&\frac{-1}{2a} - \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-1}{2a} - \frac{b}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&0 - \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})b}{2a} - \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\=&\frac{b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} - \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{b^{2}}{2(b^{2} - 4ac)^{\frac{3}{2}}a} - \frac{1}{2(b^{2} - 4ac)^{\frac{1}{2}}a}\right)}{db}\\=&\frac{(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b^{2}}{2a} + \frac{2b}{2(b^{2} - 4ac)^{\frac{3}{2}}a} - \frac{(\frac{\frac{-1}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{3}{2}}})}{2a} + 0\\=& - \frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( - \frac{3b^{3}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3b}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\right)}{db}\\=& - \frac{3(\frac{\frac{-5}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{7}{2}}})b^{3}}{2a} - \frac{3*3b^{2}}{2(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3(\frac{\frac{-3}{2}(2b + 0)}{(b^{2} - 4ac)^{\frac{5}{2}}})b}{2a} + \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\=&\frac{15b^{4}}{2(b^{2} - 4ac)^{\frac{7}{2}}a} - \frac{9b^{2}}{(b^{2} - 4ac)^{\frac{5}{2}}a} + \frac{3}{2(b^{2} - 4ac)^{\frac{3}{2}}a}\\ \end{split}\end{equation} \]