Question

导数题目:已知y=(√(1+x)-√(1-x))/(√(1+x)+√(1-x))，求y’

Answer 1

分子有理化
y=[(1+x)-(1-x)]/[√(1+x)+√(1-x)]^2
=2x/[1+x+2√(1-x^2)+1-x]
=x/[1+√(1-x^2)]

[1+√(1-x^2)]'=0+(1/2)*(1-x^2)^(-1/2)*(1-x^2)'
=-x/√(1-x^2)

所以y'={x'*[1+√(1-x^2)]-x*[1+√(1-x^2)]'}/[1+√(1-x^2)]^2
={[1+√(1-x^2)]+x^2/√(1-x^2)}/[1+√(1-x^2)]^2
={√(1-x^2)+(1-x^2)+x^2]/√(1-x^2)*[1+√(1-x^2)]^2
=[√(1-x^2)+1]/√(1-x^2)*[1+√(1-x^2)]^2
=1/√(1-x^2)*[1+√(1-x^2)]
=1/[√(1-x^2)+1-x^2]

Answer 2

先化简:
y=(√(1+x)-√(1-x))/(√(1+x)+√(1-x))(有理化)
=[根号(1+x)-根号(1-x)]^2/{[(根号(1+x)+根号(1-x)]*[(根号(1+x)-根号(1-x)]}
=[1+x+1-x-2根号(1-x^2)]/[1+x-1+x]
=[2-2根号(1-x^2)]/2x
=1/x-根号(1-x^2)/x

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