What is the derivative of the inverse of arctan of tan(x)?

[moo5003]

I don't really need help on the actual problem *I believe* I'm just confused on what its actually asking.

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...)

 

[AKG]

No. f is a function defined on (-\pi /2,\, \pi /2) by f(x) = tan(x). g is the inverse of f, and it's called arctan. So g(x) = arctan(x). This means g(f(x)) = x. In other words:

g(f(x)) = g(tan(x)) = arctan(f(x)) = arctan(tan(x)) = x

g = f^-1, g = tan^-1, arctan = f^-1, arctan = tan^-1

None of these ways of writing are more correct than the other. f and tan are the same thing. g, arctan, f^-1, tan^-1 are all the same thing.

And g(x) = arctan(x), not arctan(f(x)). arctan(f(x)) = x.

One more thing, the number \pi spelt out in English letters is "pi", not "Pie". It's a greek letter, not a baked desert. Anyways, do you understand what the question is asking you to do now?

 

[moo5003]

So basically from what I could muster from reading the book 5 times and just looking back here is that they want me to do the following:

f(x) = tan(x) = y
g(y) = arctan(y)

Arctan'(y) = 1/tan'(x) = cos^2(x)

since y = tan(x) --> cos^2(x) = 1/(y^2+1) *Took a few trig idents.

Thus Arctan'(y) = 1/(y^2 + 1)

Edit: I realize arctan(tan(x)) is x, which is why I posted to clarify the question. I don't understand what the question is asking when it says "obtain the derivative of g = arctan of f". To me that sounds like g(f(x)), though I now see they meant g(J) were J is the interval obtained with J = tan(I) were I is (-pi/2, pi/2). (J being (-1,1)).

PS: Pie is delicious, I only wish pi could be spelled pie.