Question

求微分f(x)=x(x+1)(x+2)……(x+n),则f'(0)=?设x+y=tany,则dy=

Answer 1

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)