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.

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- Mar 28th 2006, 02:19 PMtttcomraderDerivative 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. - Mar 28th 2006, 06:57 PMThePerfectHackerQuote:

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. - Mar 29th 2006, 11:01 AMalbi
One thing. The $\displaystyle x \in (-1, 1)$ assumption is unnecessary.

- Mar 29th 2006, 01:45 PMThePerfectHackerQuote:

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. - Mar 30th 2006, 05:11 AMalbiQuote:

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) - Mar 30th 2006, 01:11 PMThePerfectHackerQuote:

Originally Posted by**albi**

- Apr 1st 2006, 07:53 PMtttcomrader
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. - Apr 2nd 2006, 03:30 AMalbi
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}

$