利用y=uv,y=u'v+uv' ；.y=x^n y'=nx^(n-1)
则有
y'=(x√(1+x²))'
=x'√(1+x²)+x(√(1+x²))'
=√(1+x²)+x(1/2)(1/(√(1+x²))(1+x²)'
=√(1+x²)+x/(2√(1+x²))(2x)
=√(1+x²)+x²/√(1+x²)
=√(1+x²)+x²√(1+x²)/(1+x²)
=√(1+x²) (1+x²/(1+x²))
=√(1+x²) (1+x²+x²)/(1+x²)
=√(1+x²) (1+2x²)/(1+x²)
= (1+2x²)/√(1+x²)