y`=2x/(x²+1)
y`=[2x²+2-2x(2x)]/(x²+1)²
=(2-2x²)/(x²+1)²=0
x=±1
(-∝,-1) -1 (-1,1) 1 (1,+∝)
y`` 0 0
y=ln(x^2+1)求这个函数的凹凸区间跟拐点
y`=2x/(x²+1)
y`=[2x²+2-2x(2x)]/(x²+1)²
=(2-2x²)/(x²+1)²=0
x=±1
(-∝,-1) -1 (-1,1) 1 (1,+∝)
y`` 0 0