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

In summary, the problem is asking for the derivative of g, which is the inverse of f (arctan), where f is the tangent function and g is the inverse of f. By using Theorem 29.9, we can find the derivative of g by taking the inverse of f'(x) and substituting it into the formula (f^-1)'(y) = 1/f'(x). This means that the derivative of arctan(tan(x)) is 1/(tan(x)^2 + 1).
  • #1
moo5003
207
0
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...)
 
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  • #2
No. f is a function defined on [itex](-\pi /2,\, \pi /2)[/itex] 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 [itex]\pi[/itex] 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?
 
  • #3
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.
 
Last edited:

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

What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function with respect to its input. It measures how much a function's output changes for a small change in its input.

What is the purpose of proving the derivative?

The proof of a derivative is important because it provides a rigorous and logical explanation for why the derivative formula works. It also allows us to generalize the concept of a derivative to more complex functions.

How is the derivative formula derived?

The derivative formula is derived using a process called limit definition. This involves taking the limit of the slope of a secant line between two points on a function as the distance between the points approaches zero. The resulting limit is the formula for the derivative.

Why is the derivative important in calculus?

The derivative is important in calculus because it allows us to analyze and understand the behavior of functions. It helps us find the slope of a curve, identify maximum and minimum points, and solve optimization problems.

What is the difference between the derivative and the anti-derivative?

The derivative and anti-derivative are inverse operations. The derivative of a function gives the rate of change of that function, while the anti-derivative of a function gives the original function back. In other words, the anti-derivative is the "undoing" of the derivative.

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