# Derivative of invert tan?

• March 28th 2006, 02:19 PM
Derivative of invert tan?
I read from the rules that the derivative of (tan^-1)x is (1/x^2+1), but I cannot figure out why and I can't find any explantations on the web.

Anyone have any ideas?

Thank you.
• March 28th 2006, 06:57 PM
ThePerfectHacker
Quote:

I read from the rules that the derivative of (tan^-1)x is (1/x^2+1), but I cannot figure out why and I can't find any explantations on the web.

Anyone have any ideas?

Thank you.

Correction, you cannot find an explanation on the web EXCEPT for this site.

Anyway, since it is inverse we have, $-1
$\tan(\arctan x)=x$
Now, take the derivative of both sides, (use chain rule),
$\frac{d\arctan x}{dx}\cdot \sec^2(\arctan x)=1$.
Thus,
$\frac{d\arctan x}{dx}=\cos^2{\arctan x}$
But, $\cos^2(\arctan x)=\frac{1}{1+x^2}$
Thus,
$\frac{d\arctan x}{dx}=\frac{1}{1+x^2}$
-----------------------
Note, I assumed that the derivative of the arctangent exists. Which is true, because if a bijective function is differenciable then so is its inverse. But I did not bother with the proof here. It happens to be true.

Q.E.D.
• March 29th 2006, 11:01 AM
albi
One thing. The $x \in (-1, 1)$ assumption is unnecessary.
• March 29th 2006, 01:45 PM
ThePerfectHacker
Quote:

Originally Posted by albi
One thing. The $x \in (-1, 1)$ assumption is unnecessary.

Of course it is, the function, $f(x)=x$ is not differenciable at $x=\pm 1$ :eek:
On the interval $x\in[-1,1]$ because that is the interval for which the arc-tangent was defined. And as you know endpoints are NEVER differenciable because by definition of derivative it means that the quotient limit exists. But the problem is that this has only one sided differenciability which implies the function IS NOT differenciable.
• March 30th 2006, 05:11 AM
albi
Quote:

Originally Posted by ThePerfectHacker
Of course it is, the function, $f(x)=x$ is not differenciable at $x=\pm 1$ :eek:
On the interval $x\in[-1,1]$ because that is the interval for which the arc-tangent was defined. And as you know endpoints are NEVER differenciable because by definition of derivative it means that the quotient limit exists. But the problem is that this has only one sided differenciability which implies the function IS NOT differenciable.

No. You have missed my point. The arc-tangent is the inverse of tangent restricted to the [-pi/2; pi/2] interval. It is:

$
f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow (-\infty, +\infty)
$

$
x \mapsto f(x) = \tan x
$

and arc-tan is defined:

$
\arctan x = f^{-1}(x)
$

Cause that during the inverse the domain and counterdomain are swapped, the arc-tangent domain is (-infinity, +infinity)
• March 30th 2006, 01:11 PM
ThePerfectHacker
Quote:

Originally Posted by albi
No. You have missed my point. The arc-tangent is the inverse of tangent restricted to the [-pi/2; pi/2] interval. It is:

$
f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow (-\infty, +\infty)
$

$
x \mapsto f(x) = \tan x
$

and arc-tan is defined:

$
\arctan x = f^{-1}(x)
$

Cause that during the inverse the domain and counterdomain are swapped, the arc-tangent domain is (-infinity, +infinity)

Your right I do not know what I was thinking. But at least see how I made that mistake, I was thinking about inverse sine or cosine.
• April 1st 2006, 07:53 PM
Thanks for the solution, now I understand how the derivative of arctanx equals to cos^2(arctanx).

But would you please explain why cost^2(arctan(x)) equals to 1/(x^2+1)?

Thank you, sir.
• April 2nd 2006, 03:30 AM
albi
$
\cos^2 x + \sin^2 x = 1
$

Let $x \neq \frac{\pi}{2} + k \pi, \quad k \in \mathbb{Z}$

Dividing the equation by $\cos^2 x$ we get:

$
1 + \tan^2 x = \frac{1}{\cos^2 x}
$

Thus:
$
\cos^2 x = \frac{1}{1 + \tan^2 x}
$

Finally:
$
\cos^2 (\arctan x) = \frac{1}{1 + \tan^2 (\arctan x)} = \frac{1}{1 + x^2}
$