Can someone show me how i can prove this?

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

thanks in advance

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- Nov 18th 2009, 12:10 AMcpjproof of arccot(x)
Can someone show me how i can prove this?

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

thanks in advance - Nov 18th 2009, 12:31 AMRockHard
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 $\displaystyle tan{x^2}+1=sec{x^2}$

- Nov 18th 2009, 01:15 AMsimplependulum
If you know $\displaystyle \frac{d}{dx}( \tan^{-1}(x) ) = \frac{1}{x^2 + 1}$ but not $\displaystyle \frac{d}{dx} ( \cot^{-1}(x) ) $

Use this identity $\displaystyle \cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2} $ - Nov 18th 2009, 02:07 AMcpj
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.

- Nov 18th 2009, 03:04 AMsimplependulum

sure

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

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

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

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

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

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

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