\begin{align} (\cot\frac{1}{x})'=-\frac{1}{\tan^2\frac{1}{x}}\cdot\frac{1}{\cos^2\frac{1}{x}}\cdot(-\frac{1}{x^2})=\frac{1}{x^2\sin^2\frac{1}{x}}>0,x\in(1,+\infty) \end{align}\\
发布于 2023-07-19 17:15・IP 属地北京
\begin{align} (\cot\frac{1}{x})'=-\frac{1}{\tan^2\frac{1}{x}}\cdot\frac{1}{\cos^2\frac{1}{x}}\cdot(-\frac{1}{x^2})=\frac{1}{x^2\sin^2\frac{1}{x}}>0,x\in(1,+\infty) \end{align}\\