,(cot1x)′=−1tan21x⋅1cos21x⋅(−1x2)=1x2sin21x>0,x∈(1,+∞)\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}\\
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解法有人给了,我就配个图吧