此题是这样的吧：函数y=arctan[(1+x)/(1-x)]?若是这样,
y′=1/[1+(1+x)²/(1-x)²][(1-x)+(1+x)]/(1-x)²
=2/[(1-x)²+(1+x)²]
=2/(2+2x²)
=1/(1+x²)