本次共计算 1 个题目:每一题对 x 求 4 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数\frac{2x}{({e}^{x} - 1)} 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{2x}{({e}^{x} - 1)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{2x}{({e}^{x} - 1)}\right)}{dx}\\=&2(\frac{-(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{2}})x + \frac{2}{({e}^{x} - 1)}\\=&\frac{-2x{e}^{x}}{({e}^{x} - 1)^{2}} + \frac{2}{({e}^{x} - 1)}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-2x{e}^{x}}{({e}^{x} - 1)^{2}} + \frac{2}{({e}^{x} - 1)}\right)}{dx}\\=&-2(\frac{-2(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{3}})x{e}^{x} - \frac{2{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{({e}^{x} - 1)^{2}} + 2(\frac{-(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{2}})\\=&\frac{4x{e}^{(2x)}}{({e}^{x} - 1)^{3}} - \frac{4{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x{e}^{x}}{({e}^{x} - 1)^{2}}\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( \frac{4x{e}^{(2x)}}{({e}^{x} - 1)^{3}} - \frac{4{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x{e}^{x}}{({e}^{x} - 1)^{2}}\right)}{dx}\\=&4(\frac{-3(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{4}})x{e}^{(2x)} + \frac{4{e}^{(2x)}}{({e}^{x} - 1)^{3}} + \frac{4x({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))}{({e}^{x} - 1)^{3}} - 4(\frac{-2(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{3}}){e}^{x} - \frac{4({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{({e}^{x} - 1)^{2}} - 2(\frac{-2(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{3}})x{e}^{x} - \frac{2{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{({e}^{x} - 1)^{2}}\\=&\frac{-12x{e}^{(3x)}}{({e}^{x} - 1)^{4}} + \frac{12{e}^{(2x)}}{({e}^{x} - 1)^{3}} + \frac{12x{e}^{(2x)}}{({e}^{x} - 1)^{3}} - \frac{6{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x{e}^{x}}{({e}^{x} - 1)^{2}}\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( \frac{-12x{e}^{(3x)}}{({e}^{x} - 1)^{4}} + \frac{12{e}^{(2x)}}{({e}^{x} - 1)^{3}} + \frac{12x{e}^{(2x)}}{({e}^{x} - 1)^{3}} - \frac{6{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x{e}^{x}}{({e}^{x} - 1)^{2}}\right)}{dx}\\=&-12(\frac{-4(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{5}})x{e}^{(3x)} - \frac{12{e}^{(3x)}}{({e}^{x} - 1)^{4}} - \frac{12x({e}^{(3x)}((3)ln(e) + \frac{(3x)(0)}{(e)}))}{({e}^{x} - 1)^{4}} + 12(\frac{-3(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{4}}){e}^{(2x)} + \frac{12({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))}{({e}^{x} - 1)^{3}} + 12(\frac{-3(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{4}})x{e}^{(2x)} + \frac{12{e}^{(2x)}}{({e}^{x} - 1)^{3}} + \frac{12x({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))}{({e}^{x} - 1)^{3}} - 6(\frac{-2(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{3}}){e}^{x} - \frac{6({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{({e}^{x} - 1)^{2}} - 2(\frac{-2(({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)})) + 0)}{({e}^{x} - 1)^{3}})x{e}^{x} - \frac{2{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{({e}^{x} - 1)^{2}}\\=&\frac{48x{e}^{(4x)}}{({e}^{x} - 1)^{5}} - \frac{48{e}^{(3x)}}{({e}^{x} - 1)^{4}} - \frac{72x{e}^{(3x)}}{({e}^{x} - 1)^{4}} + \frac{48{e}^{(2x)}}{({e}^{x} - 1)^{3}} + \frac{28x{e}^{(2x)}}{({e}^{x} - 1)^{3}} - \frac{8{e}^{x}}{({e}^{x} - 1)^{2}} - \frac{2x{e}^{x}}{({e}^{x} - 1)^{2}}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!