Find the equation of the tangent to the graph y=f'(x) at x=3 where f(x)=x^4

Question

So I did :
f'(x)=4x^3
y-81=4x^3(x-3)
y=4x^4-12x+81

?? Am i even remotely correct?

Answer 1

plug in 4 for x
f'(x)=4(4)^3 =?

Answer 2

I mean plug in 3 for x
f'(x)=4(3)^3 =?

Answer 3

108

Answer 4

y-81=108(x-3)
and solve for y.

Answer 5

y=108x-243