f(x) = cos(x) + 3x

Find f^-1 (1) and (f^-1)' (1)

f^-1 (1) = 0

(f^-1)' (1) = 1/3

These are the answers given, but I also don't understand how my teacher got these.

y = cos (x)+ 3 x

interchange
x = cos y + 3 y
if y = 0, x = 1 + 0

x' = -sin y + 3
but y = 0 at x = 1
dx/dy = 3
dy/dx = 1/3