If f(x) = sin(x) +2x+1 and g is the inverse function of f, what is the value of g'(1)? Answer is 1/3. No calculator allowed. About 2 minutes allowed to solve.
g'(a)=1/f'(g(a)) (from f(g(x))=x, taking derivative of both sides)
The Attempt at a Solution
Plugging into the above equation, I just get g'(1) =1/(cos(g(1)+2), which looks like a dead end.
I thought of trying the reverse, using f'(1)=1/g'(f(1)), but that just gets me that cos(1)+2=1/g'(3), which is even further from a solution. I also thought that if I used f'(x)=1/g'(f(x)) and found x such that f(x) = 1, that would give me the answer, but then I get the equation sin(x) =-2x , which is not getting me anywhere either for a problem that is supposed to be simple.