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