Question

In Exercises 43 and 44, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. \(\frac{d}{d x}[\arctan (\tan x)]=1\) for all \(x\) in the domain.

Short answer

The statement that \(\frac{d}{d x}[\arctan (\tan x)]=1\) for all \(x\) in the domain is false. Although the derivative is 1 for \(x \in (-\pi/2, \pi/2)\), it is not 1 for all \(x\) due to periodicity of \(\tan x\).

Step-by-step solution

Compute the derivative of the given function

The derivative of \(\arctan(x)\) with respect to \(x\) is \(\frac{1}{1+x^2}\). We should apply the chain rule while taking derivative of \(\arctan (\tan x)\), which gives us \(\frac{1}{1+(\tan x)^2} \cdot \frac{d}{dx} (\tan x)\). We know that \(\frac{d}{dx} \tan x = \sec^2 x\), so the derivative of the function becomes \(\frac{\sec^2x}{1 + \tan^2x}\).

Simplify the derivative

The formula \(\tan^2x + 1 = \sec^2x\) can be applied here to simplify the derivative. This simplification yields \(\frac{\sec^2x}{\sec^2x}\), which simplifies to \(1\)

Consider the domain

We should notice that the function \(\arctan (\tan x)\) is defined over the domain \((-\pi/2, \pi/2)\). For values of \(x\) outside of this domain, \(\tan x\) simplifies to a term equivalent to some \(x + k\pi\), altering the derivative. As a result, the initial assertion that the derivative equals 1 for all \(x\) in the domain is false. It is true only for \(x \in (-\pi/2, \pi/2)\).