- #1

moo5003

- 207

- 0

Problem: "Use Theorem 29.9 to obtain the derivative of the inverse g = arctan of f where f(x) = tan(x) for x in (-Pie/2, Pie/2)"

29.9: "Let f be a 1-1 continuous function on an open interval I, and let J = f(I). If f is differentiable at x in I and if f'(x) != 0, then f^(-1) is differentiable at y = f(x) and (f^(-1))'(y) = 1/f'(x)"

I'm just trying to clarify the problem because I can't seem to grasp what its asking. They want me to find (g^(-1))' such that g(x) = arctan(f(x)) such that f(x) = tan(x)?

Ie: find the derivative of arctan(tan(x))? (Which is 1...)