令u=1+√x,∴u′=1/(2√x),u^2=1+2√x+x
∴y′
=(x/u)′=(x′u-xu′)/u^2=[u-x/(2√x)]/(1+2√x+x)
=[1+√x-x/(2√x)]/(1+x+2√x)=(2√x+2x-x)/[2√x(1+x+2√x)]
=(x+2√x)/(4x+2x√x+2√x).
令u=1+√x,∴u′=1/(2√x),u^2=1+2√x+x
∴y′
=(x/u)′=(x′u-xu′)/u^2=[u-x/(2√x)]/(1+2√x+x)
=[1+√x-x/(2√x)]/(1+x+2√x)=(2√x+2x-x)/[2√x(1+x+2√x)]
=(x+2√x)/(4x+2x√x+2√x).