求导啊
f(x)=(x^2+1)^(1/2)
f'(x)=(1/2)*(x^2+1)^(1/2-1)*(x^2+1)'
=(1/2)*(x^2+1)^(-1/2)*2x
=x/√(x^2+1)
x=1
f'(x)=1/√2
所以切线斜率等于√2/2
y-√2=√2/2*(x-1)
x-√2*y+1=0