x的x次方求导以及此类换元法进阶

一、对x的x次方求导

问题描述：对 $x^x$ 求导

解决：

设 $y=x^x$，对两边求自然对数
$$\begin{array}{rl}\ln y& =x\ln x\\ \frac{\mathrm{d}\ln y}{\mathrm{d}x}& =\frac{\mathrm{d}(x\ln x)}{\mathrm{d}x}\\ \frac{\mathrm{d}\ln y}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}& =1+\ln x\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =(1+\ln x)y\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =(1+\ln x)x^x\\ \dot{x}& =(1+\ln x)x^x\end{array}$$
可以看到这一类次方函数的求导需要构造一个 $y$ 来进行求解，我们接下来进阶

二、进阶

温馨提示，本章涉及到了大量的求导链式法则，所以还不是很懂这部分内容的小伙伴可以先去补补课，本节的内容对于想要研究高阶滑模算法的小伙伴有很大帮助，反正我是为了研究这个问题。

因为我们这章数学是为了控制服务的，这里我们直接选择一个简单二阶系统
$$\begin{array}{r}\{\begin{array}{l}{\dot{x}}_1=x_2\\ {\dot{x}}_2=u\end{array}\end{array}$$
求下面等式对时间 $t$ 导数，其中函数 $sign$ 为符号函数y
$$\begin{array}{r}y=|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}sign(x_1)\end{array}$$
先对两边同时求导解决符号函数的问题
$$\begin{array}{r}{y}^{\prime }={\left(|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\right)}^{\prime }sign(x_1)+\left(|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\right){\left(sign(x_1)\right)}^{\prime }={\left(|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\right)}^{\prime }sign(x_1)\end{array}$$
设 $z$ 为上式的第一部分,只需要对 $z$ 求导即可
$$\begin{array}{rl}z& =|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\\ \ln z& =\frac{\lambda x_1^2}{1+\mu x_1^2}\ln |x_1|\\ \frac{\mathrm{d}\ln z}{\mathrm{d}t}& =\frac{\mathrm{d}\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\ln |x_1|\right)}{\mathrm{d}t}\\ \frac{\mathrm{d}\ln z}{\mathrm{d}z}\frac{\mathrm{d}z}{\mathrm{d}t}& =\frac{\mathrm{d}\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{\mathrm{d}t}\ln |x_1|+\frac{\mathrm{d}\ln |x_1|}{\mathrm{d}t}\frac{\lambda x_1^2}{1+\mu x_1^2}\end{array}$$
等式右侧分为两部分，两部分分别求，首先是第一部分
$$\begin{array}{rl}\frac{\mathrm{d}\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{\mathrm{d}t}& =\frac{\mathrm{d}\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{\mathrm{d}{x}_1^2}\frac{\mathrm{d}{x}_1^2}{\mathrm{d}{x}_1}\frac{\mathrm{d}{x}_1}{\mathrm{d}t}\\ & =\left(\frac{\lambda (1+\mu x_1^2)-\lambda x_1^2\mu }{(1+\mu x_1^2{)}^2}\right)2x_1{\dot{x}}_1\\ & =\frac{2\lambda x_1{x}_2}{(1+\mu x_1^2{)}^2}\end{array}$$
其次是第二部分
$$\begin{array}{r}\frac{\mathrm{d}\ln |x_1|}{\mathrm{d}t}=\frac{\mathrm{d}\ln |x_1|}{\mathrm{d}|x_1|}\frac{\mathrm{d}|x_1|}{\mathrm{d}{x}_1}\frac{\mathrm{d}{x}_1}{\mathrm{d}t}=\frac{1}{|x_1|}sign(x_1)x_2=\frac{x_2}{x_1}\end{array}$$
将这两部分带入对 $z$ 的求导部分可得
$$\begin{array}{rl}\frac{\mathrm{d}\ln z}{\mathrm{d}z}\frac{\mathrm{d}z}{\mathrm{d}t}& =\frac{\mathrm{d}\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{\mathrm{d}t}\ln |x_1|+\frac{\mathrm{d}\ln |x_1|}{\mathrm{d}t}\frac{\lambda x_1^2}{1+\mu x_1^2}\\ \frac{1}{z}\frac{\mathrm{d}z}{\mathrm{d}t}& =\frac{2\lambda x_1{x}_2}{(1+\mu x_1^2{)}^2}\ln |x_1|+\frac{\lambda x_1{x}_2}{1+\mu x_1^2}\\ \frac{1}{z}\frac{\mathrm{d}z}{\mathrm{d}t}& =\frac{\lambda x_1{x}_2}{1+\mu x_1^2}\left(\frac{2\ln |x_1|}{1+\mu x_1^2}+1\right)\\ \frac{\mathrm{d}z}{\mathrm{d}t}& =\frac{\lambda x_1{x}_2}{1+\mu x_1^2}\left(\frac{2\ln |x_1|}{1+\mu x_1^2}+1\right)|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\end{array}$$
将 $z$ 带回 $y^{\prime }$ 可得
$$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}t}& =\frac{\lambda x_1{x}_2}{1+\mu x_1^2}\left(\frac{2\ln |x_1|}{1+\mu x_1^2}+1\right)|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}sign(x_1)\\ & =\frac{\lambda |x_1|x_2}{1+\mu x_1^2}\left(\frac{2\ln |x_1|}{1+\mu x_1^2}+1\right)|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\end{array}$$