来自陈中华的问题
证明/arctan(a)-arctan(b)/
证明/arctan(a)-arctan(b)/


证明/arctan(a)-arctan(b)/
证明/arctan(a)-arctan(b)/
|arctan(a)-arctan(b)|/|a-b|=arctan'(x)(求导,x在ab之间)这里用了拉格朗日中值定理
所以|arctan(a)-arctan(b)|/|a-b|=1/(1+x^2)