Answer
${\frac{d{cot^{-1}{u}}}{du}}$ = -${\frac{1}{1+u^2}}\frac{du}{dx}$
Work Step by Step
We know that:
${cot^{-1}{u}}$ = ${\frac{\pi}{2}}-{tan^{-1}{u}}$
differentiating both sides:
${\frac{d{cot^{-1}{u}}}{du}}={\frac{d\pi}{2du}}-{\frac{d{tan^{-1}{u}}}{du}}$
differentiation of constant is zero and ${\frac{d{\tan^{-1}{u}}}{du}={\frac{1}{1+u^2}}}$
thus:
${\frac{d{cot^{-1}{u}}}{du}}$ = -${\frac{1}{1+u^2}}\frac{du}{dx}$
