y=arth(thx)的导数y'=[arth(thx)]'(thx)'={1/[1-th^2(x)]}*[1/ch^2(x)]=1/[ch^2(x)-sh^2(x)]=1请问求导高手们,为什么答案不对?答案是1/[1+2sh^2(x)]

Question

y=arth(thx)的导数
y'=[arth(thx)]'(thx)'={1/[1-th^2(x)]}*[1/ch^2(x)]=1/[ch^2(x)-sh^2(x)]=1
请问求导高手们,
为什么答案不对?
答案是1/[1+2sh^2(x)]

Answer 1

公式：(arthu)'=u'/(1-u^2) ( |u|