原題：求 y=arctan[(1+x)²/(1-x)²] 的導(dǎo)數(shù)
將要用到的求導(dǎo)公式：
y=arctanxy'=1/(1+x²)
兩函數(shù)的商求導(dǎo)： [u(x)/v(x)]'=[u'(x)v(x)-u(x)v'(x)]/v²(x)
y=(1+x)²y'=2+2x
y=(1-x)²y'=-2+2x
根據(jù)復(fù)合函數(shù)求導(dǎo)法則,有
y'=1/[1+(1+x)^4/(1-x)^4] * [(1+x)²/(1-x)²]'
=1/[1+(1+x)^4/(1-x)^4] * [(2+2x)(1-x)²-(1+x)²(-2+2x)]/(1-x)^4
=1/[(1-x)^4+(1+x)^4] * [(2+2x)(1-x)²-(1+x)²(-2+2x)]
=1/[(1-x)^4+(1+x)^4] * 2[(1+x)(1-x)²+(1+x)²(1-x)]
=1/[(1-x)^4+(1+x)^4] * 2(1-x²)*2
=4(1-x²)/[(1-x)^4+(1+x)^4]
=4(1-x²)/[2(1+6x²+x^4)]
=2(1-x²)/(1+6x²+x^4)