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.Originally Posted by tttcomrader
Anyway, since it is inverse we have, $\displaystyle -1<x<1$
$\displaystyle \tan(\arctan x)=x$
Now, take the derivative of both sides, (use chain rule),
$\displaystyle \frac{d\arctan x}{dx}\cdot \sec^2(\arctan x)=1$.
Thus,
$\displaystyle \frac{d\arctan x}{dx}=\cos^2{\arctan x}$
But, $\displaystyle \cos^2(\arctan x)=\frac{1}{1+x^2}$
Thus,
$\displaystyle \frac{d\arctan x}{dx}=\frac{1}{1+x^2}$
-----------------------
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.
Of course it is, the function, $\displaystyle f(x)=x$ is not differenciable at $\displaystyle x=\pm 1$Originally Posted by albi
On the interval $\displaystyle x\in[-1,1]$ 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:Originally Posted by ThePerfectHacker
$\displaystyle
f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow (-\infty, +\infty)
$
$\displaystyle
x \mapsto f(x) = \tan x
$
and arc-tan is defined:
$\displaystyle
\arctan x = f^{-1}(x)
$
Cause that during the inverse the domain and counterdomain are swapped, the arc-tangent domain is (-infinity, +infinity)
Start with:
$\displaystyle
\cos^2 x + \sin^2 x = 1
$
Let $\displaystyle x \neq \frac{\pi}{2} + k \pi, \quad k \in \mathbb{Z}$
Dividing the equation by $\displaystyle \cos^2 x $ we get:
$\displaystyle
1 + \tan^2 x = \frac{1}{\cos^2 x}
$
Thus:
$\displaystyle
\cos^2 x = \frac{1}{1 + \tan^2 x}
$
Finally:
$\displaystyle
\cos^2 (\arctan x) = \frac{1}{1 + \tan^2 (\arctan x)} = \frac{1}{1 + x^2}
$