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。\[ \begin{equation}\begin{split}【1/1】求函数a + bx - {e^{2}}^{x}((f)cos(x) + (g)sin(x)) 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = a + bx - f{e^{2}}^{x}cos(x) - g{e^{2}}^{x}sin(x)\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( a + bx - f{e^{2}}^{x}cos(x) - g{e^{2}}^{x}sin(x)\right)}{dx}\\=&0 + b - f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))cos(x) - f{e^{2}}^{x}*-sin(x) - g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))sin(x) - g{e^{2}}^{x}cos(x)\\=&b - f{e^{2}}^{x}ln(e^{2})cos(x) + f{e^{2}}^{x}sin(x) - g{e^{2}}^{x}ln(e^{2})sin(x) - g{e^{2}}^{x}cos(x)\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( b - f{e^{2}}^{x}ln(e^{2})cos(x) + f{e^{2}}^{x}sin(x) - g{e^{2}}^{x}ln(e^{2})sin(x) - g{e^{2}}^{x}cos(x)\right)}{dx}\\=&0 - f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln(e^{2})cos(x) - \frac{f{e^{2}}^{x}e^{2}*0cos(x)}{(e^{2})} - f{e^{2}}^{x}ln(e^{2})*-sin(x) + f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))sin(x) + f{e^{2}}^{x}cos(x) - g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln(e^{2})sin(x) - \frac{g{e^{2}}^{x}e^{2}*0sin(x)}{(e^{2})} - g{e^{2}}^{x}ln(e^{2})cos(x) - g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))cos(x) - g{e^{2}}^{x}*-sin(x)\\=& - f{e^{2}}^{x}ln^{2}(e^{2})cos(x) + 2f{e^{2}}^{x}ln(e^{2})sin(x) + f{e^{2}}^{x}cos(x) - g{e^{2}}^{x}ln^{2}(e^{2})sin(x) + g{e^{2}}^{x}sin(x) - 2g{e^{2}}^{x}ln(e^{2})cos(x)\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( - f{e^{2}}^{x}ln^{2}(e^{2})cos(x) + 2f{e^{2}}^{x}ln(e^{2})sin(x) + f{e^{2}}^{x}cos(x) - g{e^{2}}^{x}ln^{2}(e^{2})sin(x) + g{e^{2}}^{x}sin(x) - 2g{e^{2}}^{x}ln(e^{2})cos(x)\right)}{dx}\\=& - f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln^{2}(e^{2})cos(x) - \frac{f{e^{2}}^{x}*2ln(e^{2})e^{2}*0cos(x)}{(e^{2})} - f{e^{2}}^{x}ln^{2}(e^{2})*-sin(x) + 2f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln(e^{2})sin(x) + \frac{2f{e^{2}}^{x}e^{2}*0sin(x)}{(e^{2})} + 2f{e^{2}}^{x}ln(e^{2})cos(x) + f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))cos(x) + f{e^{2}}^{x}*-sin(x) - g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln^{2}(e^{2})sin(x) - \frac{g{e^{2}}^{x}*2ln(e^{2})e^{2}*0sin(x)}{(e^{2})} - g{e^{2}}^{x}ln^{2}(e^{2})cos(x) + g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))sin(x) + g{e^{2}}^{x}cos(x) - 2g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln(e^{2})cos(x) - \frac{2g{e^{2}}^{x}e^{2}*0cos(x)}{(e^{2})} - 2g{e^{2}}^{x}ln(e^{2})*-sin(x)\\=& - f{e^{2}}^{x}ln^{3}(e^{2})cos(x) + 3f{e^{2}}^{x}ln^{2}(e^{2})sin(x) - f{e^{2}}^{x}sin(x) + 3f{e^{2}}^{x}ln(e^{2})cos(x) - g{e^{2}}^{x}ln^{3}(e^{2})sin(x) + 3g{e^{2}}^{x}ln(e^{2})sin(x) - 3g{e^{2}}^{x}ln^{2}(e^{2})cos(x) + g{e^{2}}^{x}cos(x)\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( - f{e^{2}}^{x}ln^{3}(e^{2})cos(x) + 3f{e^{2}}^{x}ln^{2}(e^{2})sin(x) - f{e^{2}}^{x}sin(x) + 3f{e^{2}}^{x}ln(e^{2})cos(x) - g{e^{2}}^{x}ln^{3}(e^{2})sin(x) + 3g{e^{2}}^{x}ln(e^{2})sin(x) - 3g{e^{2}}^{x}ln^{2}(e^{2})cos(x) + g{e^{2}}^{x}cos(x)\right)}{dx}\\=& - f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln^{3}(e^{2})cos(x) - \frac{f{e^{2}}^{x}*3ln^{2}(e^{2})e^{2}*0cos(x)}{(e^{2})} - f{e^{2}}^{x}ln^{3}(e^{2})*-sin(x) + 3f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln^{2}(e^{2})sin(x) + \frac{3f{e^{2}}^{x}*2ln(e^{2})e^{2}*0sin(x)}{(e^{2})} + 3f{e^{2}}^{x}ln^{2}(e^{2})cos(x) - f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))sin(x) - f{e^{2}}^{x}cos(x) + 3f({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln(e^{2})cos(x) + \frac{3f{e^{2}}^{x}e^{2}*0cos(x)}{(e^{2})} + 3f{e^{2}}^{x}ln(e^{2})*-sin(x) - g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln^{3}(e^{2})sin(x) - \frac{g{e^{2}}^{x}*3ln^{2}(e^{2})e^{2}*0sin(x)}{(e^{2})} - g{e^{2}}^{x}ln^{3}(e^{2})cos(x) + 3g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln(e^{2})sin(x) + \frac{3g{e^{2}}^{x}e^{2}*0sin(x)}{(e^{2})} + 3g{e^{2}}^{x}ln(e^{2})cos(x) - 3g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))ln^{2}(e^{2})cos(x) - \frac{3g{e^{2}}^{x}*2ln(e^{2})e^{2}*0cos(x)}{(e^{2})} - 3g{e^{2}}^{x}ln^{2}(e^{2})*-sin(x) + g({e^{2}}^{x}((1)ln(e^{2}) + \frac{(x)(e^{2}*0)}{(e^{2})}))cos(x) + g{e^{2}}^{x}*-sin(x)\\=& - f{e^{2}}^{x}ln^{4}(e^{2})cos(x) + 4f{e^{2}}^{x}ln^{3}(e^{2})sin(x) - 4f{e^{2}}^{x}ln(e^{2})sin(x) + 6f{e^{2}}^{x}ln^{2}(e^{2})cos(x) - f{e^{2}}^{x}cos(x) - g{e^{2}}^{x}ln^{4}(e^{2})sin(x) + 6g{e^{2}}^{x}ln^{2}(e^{2})sin(x) - 4g{e^{2}}^{x}ln^{3}(e^{2})cos(x) - g{e^{2}}^{x}sin(x) + 4g{e^{2}}^{x}ln(e^{2})cos(x)\\ \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!