proof of arccot(x)

**cpj** · November 18th 2009, 12:10 AM

Can someone show me how i can prove this?

(d/dx)cot^-1 =-1/1+(x^2)

thanks in advance

**RockHard** · November 18th 2009, 12:31 AM

I would like to know this as well, would doing the integral of his solution be a way to provide some sort of "proof" by using the identity $tan{x^2}+1=sec{x^2}$

**simplependulum** · November 18th 2009, 01:15 AM

If you know $\frac{d}{dx}( \tan^{-1}(x) ) = \frac{1}{x^2 + 1}$ but not $\frac{d}{dx} ( \cot^{-1}(x) )$

Use this identity $\cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2}$

**cpj** · November 18th 2009, 02:07 AM

thanks, but would it be possible for you to elaborate, maybe show some working, i'm not sure how to use that identity to help with the proof.

**simplependulum** · November 18th 2009, 03:04 AM

Originally Posted by simplependulum

If you know $\frac{d}{dx}( \tan^{-1}(x) ) = \frac{1}{x^2 + 1}$ but not $\frac{d}{dx} ( \cot^{-1}(x) )$

Use this identity $\cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2}$

sure

We have $\cot(\theta) = \frac{1}{\tan(\theta)} = \tan( \frac{\pi}{2} - \theta )$

$\theta = \cot^{-1}( \tan( \frac{\pi}{2} - \theta ))$ -(1)

Let $x = \tan( \frac{\pi}{2} - \theta )$

$\tan^{-1}(x) = \frac{\pi}{2} - \theta$ -(2)

$(2) - (1)$ , it is to eliminate $\theta$

$\tan^{-1}(x) = \frac{\pi}{2} - \cot^{-1}( \tan( \frac{\pi}{2} - \theta ))$

$\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$

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