Derivation of x to the power of x and advancement of this type of substitution method

1. Derivation of x to the power of x

Problem description: Yes $x^{}$Derivative of${}^x$

solve:

设$y=x^x$，对两边求自然对数
$$\begin{array}{rl}\ln y& =x\ln x\\ \frac{\mathrm{d}\ln y}{\mathrm{d}x}& =\frac{\mathrm{d}(x\ln x)}{dx}\\ \frac{d\ln y}{\mathrm{d}y}\frac{\mathrm{d}y}{dx}& =1+\ln x\\ \frac{dy}{dx}& =(1+\ln x)y\\ \frac{dy}{dx}& =(1+\ln x)x^x\\ \dot{x}& =(1+\ln x)x^x\end{array}$$
It can be seen that the derivation of this type of power function requires the construction of a $y$ to solve, we will advance

2. Advanced

Warm reminder, this chapter involves a lot of derivation chain rules, so friends who don’t understand this part of the content can go and make up for it first. The content of this section is very useful for friends who want to study high-order sliding mode algorithms. Help, I'm here to research this problem anyway.

Because our mathematics in this chapter is for control services, here we directly choose a simple second-order system
$$\begin{array}{r}\{\begin{array}{l}{\dot{x}}_1=x_2\\ {\dot{x}}_2=u\end{array}\end{array}$$
Find the following equation for time $t$ derivative, where the function$sign$ is the sign function y
$$\begin{array}{r}y=|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}sign(x_1)\end{array}$$
First, solve the problem of the sign function by derivation of both sides at the same time
$$\begin{array}{r}{y}^{\prime }={\left(|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\right)}^{\prime }sign(x_1)+\left(|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\right){\left(sign(x_1)\right)}^{\prime }={\left(|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\right)}^{\prime }sign(x_1)\end{array}$$
readzz $z$ is the first part of the above formula, we only need to compare$z$ 求导即可
$$\begin{array}{rl}z& =|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\\ \ln z& =\frac{\lambda x_1^2}{1+\mu x{\_}_1^2}\ln |x_1|\\ \frac{\mathrm{d}\ln z}{\mathrm{d}t}& =\frac{d\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\ln |x_1|\right)}{dt}\\ \frac{\mathrm{d}\ln z}{\mathrm{d}z}\frac{\mathrm{d}z}{\mathrm{d}t}& =\frac{d\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{dt}\ln |x_1|+\frac{d\ln |x_1|}{dt}\frac{\lambda x_1^2}{1+\mu x{\_}_1^2}\end{array}$$
The right side of the equation is divided into two parts, and the two parts are calculated separately. The first part is
$$\begin{array}{rl}\frac{\mathrm{d}\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{\mathrm{d}t}& =\frac{d\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{\mathrm{d}{x}_1^2}\frac{d{x}_1^2}{d{x}_1}\frac{d{x}_1}{dt}\\ & =\left(\frac{l(1+\mu x{\_}_1^2)-\lambda x_1^2}{(1+\mu x{\_}_1^2{)}^2}\right)2x{\_}_1{\dot{x}}_1\\ & =\frac{2l{x}_1{x}_2}{(1+\mu x{\_}_1^2{)}^2}\end{array}$$
其次是第二部分
$$\begin{array}{r}\frac{\mathrm{d}\ln |x_1|}{\mathrm{d}t}=\frac{d\ln |x_1|}{\mathrm{d}|x_1|}\frac{d|x_1|}{\mathrm{d}{x}_1}\frac{d{x}_1}{dt}=\frac{1}{|x_1|}sign(x_1)x_2=\frac{x_2}{x_1}\end{array}$$
Bring these two parts into the zzThe derivative part of$z\; can\; be\; obtained$
$$\begin{array}{rl}\frac{\mathrm{d}\ln z}{\mathrm{d}z}\frac{\mathrm{d}z}{\mathrm{d}t}& =\frac{d\left(\frac{\lambda x_1^2}{1+\mu x_1^2}\right)}{dt}\ln |x_1|+\frac{d\ln |x_1|}{dt}\frac{\lambda x_1^2}{1+\mu x{\_}_1^2}\\ \frac{1}{z}\frac{dz}{dt}& =\frac{2l{x}_1{x}_2}{(1+\mu x{\_}_1^2{)}^2}\ln |x_1|+\frac{\lambda x_1{x}_2}{1+\mu x{\_}_1^2}\\ \frac{1}{z}\frac{dz}{dt}& =\frac{\lambda x_1{x}_2}{1+\mu x{\_}_1^2}\left(\frac{2\ln |x_1|}{1+\mu x{\_}_1^2}+1\right)\\ \frac{dz}{dt}& =\frac{\lambda x_1{x}_2}{1+\mu x{\_}_1^2}\left(\frac{2\ln |x_1|}{1+\mu x{\_}_1^2}+1\right)|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\end{array}$$
General $z$ brings back$y^{\prime }$ 可得
$$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}t}& =\frac{\lambda x_1{x}_2}{1+\mu x{\_}_1^2}\left(\frac{2\ln |x_1|}{1+\mu x{\_}_1^2}+1\right)|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}sign(x_1)\\ & =\frac{\lambda |x_1|x_2}{1+\mu x{\_}_1^2}\left(\frac{2\ln |x_1|}{1+\mu x{\_}_1^2}+1\right)|x_1{|}^{\frac{\lambda x_1^2}{1+\mu x_1^2}}\end{array}$$