Question

y=x/根号下x^2+1 求导如题

Answer 1

(u/v)'=(u'*v-u*v')/v²

这里u=x,v=√(x²+1)=(x²+1)^(1/2)

u'=1

v'=1/2*(x²+1)^(1/2-1)*(2x)'

=x/√(x²+1)

所以y'=[1*√(x²+1)-x*x/√(x²+1)]/(x²+1)

=[(x²+1-x²)/√(x²+1)]/(x²+1)

=1/[(x²+1)√(x²+1)]