Question

函数f(x)=根号下（1+X方）在点（1,√2）的切线的斜率

Answer 1

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