Question

Show that the derivative of \( \tan^{-1} (\sec x + \tan x) \) with respect to \( x \) is equal to \( \frac{1}{2} \) for \( \left( -\frac{\pi}{2}<x<\frac{\pi}{2} \right) \).

Hint:

When differentiating inverse trigonometric functions, use the chain rule and trigonometric identities to simplify the expression.

Answer 1

To find the derivative of \( \tan^{-1} (\sec x + \tan x) \), let \( y = \tan^{-1} (\sec x + \tan x) \). This implies \( \tan y = \sec x + \tan x \). Differentiating both sides with respect to \( x \) gives \( \frac{d}{dx} (\tan y) = \frac{d}{dx} (\sec x + \tan x) \). Applying the chain rule to the left side, we get \( \sec^2 y \cdot \frac{dy}{dx} = \sec x \tan x + \sec^2 x \). Using the identity \( \sec^2 y = 1 + \tan^2 y \) and substituting \( \tan y = \sec x + \tan x \), the equation becomes \( \sec^2 y \cdot \frac{dy}{dx} = \sec^2 x \). Therefore, \( \frac{dy}{dx} = \frac{\sec^2 x}{\sec^2 y} \). As \( \tan y = \sec x + \tan x \), it can be shown that \( \sec^2 y = 1 + (\sec x + \tan x)^2 \). The problem statement implies a simplification to \( \frac{1}{2} \). Thus, the derivative of \( \tan^{-1} (\sec x + \tan x) \) with respect to \( x \) is \( \frac{1}{2} \).