Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 423: 118

$\displaystyle \frac{d}{dx}[\cot^{-1}u]=-\frac{1}{1+u^{2}}\cdot\frac{du}{dx}$

$\displaystyle \cot^{-1}u=\frac{\pi}{2}-\tan^{-1}u\qquad$ We take the derivative of both sides: $\displaystyle \frac{d}{dx}[\cot^{-1}u]= 0- [\displaystyle \frac{1}{1+u^{2}}]\frac{du}{dx}$ $\displaystyle \frac{d}{dx}[\cot^{-1}u]=-\frac{1}{1+u^{2}}\cdot\frac{du}{dx}$