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It supports mathematical functions (including trigonometric functions).
Enter the mathematical formula directly and click the "Next" button to get the calculation answer.
It supports mathematical functions (including trigonometric functions).
Current location：Mathematical operation > History of Mathematical Computation > Answer
Overview: 1 questions will be solved this time.Among them
☆1 integer calculations
[1/1 Integer column vertical calculation]
Question type: Integer multiplication
Original question: 113*8849557522123893805309734513274660498185840707964601769911504424778761061946902654867256637168141p;Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ ln(1 + {e}^{(2x)}) + \frac{ln(1 + {e}^{(-2x)})}{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = ln({e}^{(2x)} + 1) + \frac{1}{2}ln({e}^{(-2x)} + 1)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( ln({e}^{(2x)} + 1) + \frac{1}{2}ln({e}^{(-2x)} + 1)\right)}{dx}\\=&\frac{(({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)})) + 0)}{({e}^{(2x)} + 1)} + \frac{\frac{1}{2}(({e}^{(-2x)}((-2)ln(e) + \frac{(-2x)(0)}{(e)})) + 0)}{({e}^{(-2x)} + 1)}\\=&\frac{2{e}^{(2x)}}{({e}^{(2x)} + 1)} - \frac{{e}^{(-2x)}}{({e}^{(-2x)} + 1)}\\ \end{split}\end{equation} \]
☆1 integer calculations
[1/1 Integer column vertical calculation]
Question type: Integer multiplication
Original question: 113*8849557522123893805309734513274660498185840707964601769911504424778761061946902654867256637168141p;Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ ln(1 + {e}^{(2x)}) + \frac{ln(1 + {e}^{(-2x)})}{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = ln({e}^{(2x)} + 1) + \frac{1}{2}ln({e}^{(-2x)} + 1)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( ln({e}^{(2x)} + 1) + \frac{1}{2}ln({e}^{(-2x)} + 1)\right)}{dx}\\=&\frac{(({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)})) + 0)}{({e}^{(2x)} + 1)} + \frac{\frac{1}{2}(({e}^{(-2x)}((-2)ln(e) + \frac{(-2x)(0)}{(e)})) + 0)}{({e}^{(-2x)} + 1)}\\=&\frac{2{e}^{(2x)}}{({e}^{(2x)} + 1)} - \frac{{e}^{(-2x)}}{({e}^{(-2x)} + 1)}\\ \end{split}\end{equation} \]