Chapter 3.3, Problem 63ES

(a) Use Formula (2) to prove that

${\left.\frac{d}{dx}\left[{\cot }^{-1}x\right]\right|}_{x=0}=-1$

(b) Use part (a) above, part (a) of Exercise $30$ in Section $1.7$ , and the chain rule to show that

$\frac{d}{dx}\left[{\cot }^{-1}x\right]=-\frac{1}{1+x^2}$

for $-\infty <x<+\infty$ .

(c) Conclude from part (b) that

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

for $-\infty <u<+\infty$ .