是y=arctan [(1-x)/(1+x)]么? 那么 y'= 1/ [1+(1-x)²/(1+x)²] * [(1-x)/(1+x)]' = (1+x)²/[(1+x)²+(1-x)²] * [-1+ 2/(1+x)]' = (1+x)²/(2+2x²) * -2/(1+x)² = -1/(1+x²)
