Question

y=x/根号下x^2+1 求导

如题

Answer 1

y=x/根号下x^2+1

y'=[x'*根号(x^2+1)-x*(根号x^2+1)']/(x^2+1)
=[根号(x^2+1)-x*(x^2+1)'/(2根号x^2+1)]/(x^2+1)
=[根号(x^2+1)-x*2x/(2根号x^2+1)]/(x^2+1)
=[根号(x^2+1)-x^2/根号(x^2+1)]/(x^2+1)
=(x^2+1-x^2)/[根号(x^2+1)*(x^2+1)]
=(x^2+1)^(-3/2)

Answer 2

(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)]