User:John Wormell/frank

$$\frac{d}{dx} \tan^{-1}(\cot x) = \frac{d}{dx}\cot x \cdot \frac{d}{d\cot x}\tan^{-1}(\cot x)$$

Let $$u = \cot x$$.

$$\frac{d}{dx}\cot x \cdot \frac{d}{du}\tan^{-1} u = -\csc^{2} x \cdot \frac{1}{1 + u^{2}} = -\csc^{2} x \cdot \frac{1}{1 + \cot^{2} x} = - \csc^{2} x \cdot \frac{1}{\csc^{2} x} = -1 $$