Question

y=1/根号x^2+1求导要明细过程,不是要答案

Answer 1

u=x²+1

则u'=2x

v=√u

则v'=1/(2√u)*u'

y=1/v

所以y'=-1/v²*v'

=-1/(x²+1)*1/[2√(x²+1)]*u'

=-1/(x²+1)*1/[2√(x²+1)]*2x

=-x/[(x²+1)√(x²+1)]