y=x/(x²+1) 求二阶导数
解：y′=[(x²+1)x′-x(x²+1)′ ]/(x²+1)²=(x²+1-2x²)/(x²+1)²=(1-x²)/(x²+1)²
y″={(x²+1)²(1-x²)′-(1-x²)[(x²+1)²]′}/(x²+1)⁴=[-2x(x²+1)²-2x(1-x²)]/(x²+1)⁴=-2x³/(x²+1)³
y=x/(x^2+1)
y'=(x^2+1-x*2x)/(x^2+1)^2=(-x^2+1)/(x^2+1)^2
y''=[-2x(x^2+1)^2+(-x^2+1)*2(x^2+1)*2x]/(x^2+1)^4
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