Can someone show me how i can prove this?
(d/dx)cot^-1 =-1/1+(x^2)
thanks in advance
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 )) $
$\displaystyle \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$