Chapter 4.3, Problem 33E

To determine

To calculate: The derivative of the given function.

Answer to Problem 33E

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

Explanation of Solution

Given information: The functionis given as,

  $\mathrm{arccot}$

Formula Used: ${\cot }^{-1}\left(x\right)=\frac{\pi }{2}-{\tan }^{-1}\left(x\right)$

The inverse function identity shown can be used to calculate the derivative of given function.

Calculation: Put given function as a function of x ,

  $\begin{array}{l}\mathrm{arccot}\left(x\right)\\ \text{Use}\text{identity,}\\ {\cot }^{-1}\left(x\right)=\frac{\pi }{2}-{\tan }^{-1}\left(x\right)\\ \text{differentiate}\text{both}\text{sides,}\\ \frac{d\left({\cot }^{-1}\left(x\right)\right)}{dx}=\frac{d\left(\frac{\pi }{2}-{\tan }^{-1}\left(x\right)\right)}{dx}\\ \frac{d\left({\cot }^{-1}\left(x\right)\right)}{dx}=0-\frac{1}{1+x^2}\\ \frac{d\left({\cot }^{-1}\left(x\right)\right)}{dx}=\frac{-1}{1+x^2}\end{array}$

So the derivative of the given functionis $\frac{d\left({\cot }^{-1}\left(x\right)\right)}{dx}=\frac{-1}{1+x^2}$ .