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current location:Derivative function > Derivative function calculation history > Answer
    There are 1 questions in this calculation: for each question, the 4 derivative of d is calculated.
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\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ \frac{d({d}^{d})}{(dd)}\ with\ respect\ to\ d:\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{{d}^{d}}{d}\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( \frac{{d}^{d}}{d}\right)}{dd}\\=&\frac{-{d}^{d}}{d^{2}} + \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d}\\=&\frac{{d}^{d}ln(d)}{d} - \frac{{d}^{d}}{d^{2}} + \frac{{d}^{d}}{d}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{{d}^{d}ln(d)}{d} - \frac{{d}^{d}}{d^{2}} + \frac{{d}^{d}}{d}\right)}{dd}\\=&\frac{-{d}^{d}ln(d)}{d^{2}} + \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln(d)}{d} + \frac{{d}^{d}}{d(d)} - \frac{-2{d}^{d}}{d^{3}} - \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d^{2}} + \frac{-{d}^{d}}{d^{2}} + \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d}\\=&\frac{-2{d}^{d}ln(d)}{d^{2}} + \frac{{d}^{d}ln^{2}(d)}{d} + \frac{2{d}^{d}ln(d)}{d} + \frac{2{d}^{d}}{d^{3}} - \frac{{d}^{d}}{d^{2}} + \frac{{d}^{d}}{d}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2{d}^{d}ln(d)}{d^{2}} + \frac{{d}^{d}ln^{2}(d)}{d} + \frac{2{d}^{d}ln(d)}{d} + \frac{2{d}^{d}}{d^{3}} - \frac{{d}^{d}}{d^{2}} + \frac{{d}^{d}}{d}\right)}{dd}\\=&\frac{-2*-2{d}^{d}ln(d)}{d^{3}} - \frac{2({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln(d)}{d^{2}} - \frac{2{d}^{d}}{d^{2}(d)} + \frac{-{d}^{d}ln^{2}(d)}{d^{2}} + \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln^{2}(d)}{d} + \frac{{d}^{d}*2ln(d)}{d(d)} + \frac{2*-{d}^{d}ln(d)}{d^{2}} + \frac{2({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln(d)}{d} + \frac{2{d}^{d}}{d(d)} + \frac{2*-3{d}^{d}}{d^{4}} + \frac{2({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d^{3}} - \frac{-2{d}^{d}}{d^{3}} - \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d^{2}} + \frac{-{d}^{d}}{d^{2}} + \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d}\\=&\frac{6{d}^{d}ln(d)}{d^{3}} - \frac{3{d}^{d}ln^{2}(d)}{d^{2}} + \frac{{d}^{d}ln^{3}(d)}{d} + \frac{3{d}^{d}ln^{2}(d)}{d} - \frac{3{d}^{d}ln(d)}{d^{2}} + \frac{3{d}^{d}ln(d)}{d} - \frac{6{d}^{d}}{d^{4}} + \frac{2{d}^{d}}{d^{3}} + \frac{{d}^{d}}{d}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{6{d}^{d}ln(d)}{d^{3}} - \frac{3{d}^{d}ln^{2}(d)}{d^{2}} + \frac{{d}^{d}ln^{3}(d)}{d} + \frac{3{d}^{d}ln^{2}(d)}{d} - \frac{3{d}^{d}ln(d)}{d^{2}} + \frac{3{d}^{d}ln(d)}{d} - \frac{6{d}^{d}}{d^{4}} + \frac{2{d}^{d}}{d^{3}} + \frac{{d}^{d}}{d}\right)}{dd}\\=&\frac{6*-3{d}^{d}ln(d)}{d^{4}} + \frac{6({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln(d)}{d^{3}} + \frac{6{d}^{d}}{d^{3}(d)} - \frac{3*-2{d}^{d}ln^{2}(d)}{d^{3}} - \frac{3({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln^{2}(d)}{d^{2}} - \frac{3{d}^{d}*2ln(d)}{d^{2}(d)} + \frac{-{d}^{d}ln^{3}(d)}{d^{2}} + \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln^{3}(d)}{d} + \frac{{d}^{d}*3ln^{2}(d)}{d(d)} + \frac{3*-{d}^{d}ln^{2}(d)}{d^{2}} + \frac{3({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln^{2}(d)}{d} + \frac{3{d}^{d}*2ln(d)}{d(d)} - \frac{3*-2{d}^{d}ln(d)}{d^{3}} - \frac{3({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln(d)}{d^{2}} - \frac{3{d}^{d}}{d^{2}(d)} + \frac{3*-{d}^{d}ln(d)}{d^{2}} + \frac{3({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))ln(d)}{d} + \frac{3{d}^{d}}{d(d)} - \frac{6*-4{d}^{d}}{d^{5}} - \frac{6({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d^{4}} + \frac{2*-3{d}^{d}}{d^{4}} + \frac{2({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d^{3}} + \frac{-{d}^{d}}{d^{2}} + \frac{({d}^{d}((1)ln(d) + \frac{(d)(1)}{(d)}))}{d}\\=&\frac{-24{d}^{d}ln(d)}{d^{4}} + \frac{12{d}^{d}ln^{2}(d)}{d^{3}} - \frac{4{d}^{d}ln^{3}(d)}{d^{2}} + \frac{{d}^{d}ln^{4}(d)}{d} + \frac{4{d}^{d}ln^{3}(d)}{d} - \frac{6{d}^{d}ln^{2}(d)}{d^{2}} + \frac{6{d}^{d}ln^{2}(d)}{d} + \frac{8{d}^{d}ln(d)}{d^{3}} + \frac{4{d}^{d}ln(d)}{d} - \frac{{d}^{d}}{d^{3}} + \frac{2{d}^{d}}{d^{2}} + \frac{24{d}^{d}}{d^{5}} - \frac{6{d}^{d}}{d^{4}} + \frac{{d}^{d}}{d}\\ \end{split}\end{equation} \]





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