quick inverse question

[kettles]

hey guys i desperately need help with this question

Given f(x)=x^3+x^2+x+1 and f-1(x) is the inverse function of y=f(x). Find the equation of the tangent to the curve y=f-1(x) at the point (4,1)

[brightsky]

Okay we need to find dy/dx first:

dy/dx = 3x^2 + 2x + 1
This means dx/dy = 1/(3x^2 + 2x + 1)
which is the derivative of the inverse function
At the point (4,1):
dx/dy = 1/(3*4^2 + 2*4 + 1) = 1/57
Using gradient point formula, we have:
y - 1 = 1/57(x - 4)

Simplify that as applicable.

[evaever]

Okay we need to find dy/dx first:

dy/dx = 3x^2 + 2x + 1
This means dx/dy = 1/(3x^2 + 2x + 1)
which is the derivative of the inverse function
At the point (4,1):
dx/dy = 1/(3*4^2 + 2*4 + 1) = 1/57
Using gradient point formula, we have:
y - 1 = 1/57(x - 4)

Simplify that as applicable.

isnt it y=x/6 +1/3 ?