calculus problem?

Question

f(x)=x^3+x, g(x)=f^-1(x), g(2)=1

what is the value of g'(2)?
i am not sure how to do this, so please help me

Answer 1

f(x)=x^3+x, g(x)=f^-1(x), g(2)=1 => f^-1(2)=1

y = x^3 + x
1 = 3x^2 (dx/dy) + (dx/dy)

When x=2,
1 = 3(2^2) (dx/dy) + (dx/dy)
1 = 13 (dx/dy)
dy/dx = 13
3x^2 + 1 = 13
3x^2 = 12
x^2 = 4
x = 2, -2
g'(2) = 2, -2

Answer 2

f´(x) = 3x^2+1
g´(x) = 1/ (3x^2+1) but x= 2 for the inverse function corresponds to x= 1 for f(x)
so g´(2) = 1/4