# Thread: proof of arccot(x)

1. ## proof of arccot(x)

Can someone show me how i can prove this?

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

thanks in advance

2. 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 $tan{x^2}+1=sec{x^2}$

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

4. 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.

5. 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 $\theta$

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

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