## University Calculus: Early Transcendentals (3rd Edition)

${\frac{d{cot^{-1}{u}}}{du}}$ = -${\frac{1}{1+u^2}}\frac{du}{dx}$
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: ${\frac{d{cot^{-1}{u}}}{du}}$ = -${\frac{1}{1+u^2}}\frac{du}{dx}$