f'(x)
={x[(x+1)(x+2)……(x+n)]}'
=(x)'[(x+1)(x+2)……(x+n)]+x[(x+1)(x+2)……(x+n)]'
=[(x+1)(x+2)……(x+n)]+x[(x+1)(x+2)……(x+n)]'
令P(x)=[(x+1)(x+2)……(x+n)]'
f'(0)=1*2*..*n+0*P(0)=n!
x+y=tany
两边同时求微分dx+dy=dy/cos(y^2)
解得dy=1/tan(y^2)