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

We need to find the derivative of \( \tan^{-1} (\sec x + \tan x) \). Let \( y = \tan^{-1} (\sec x + \tan x) \), so: \[ \tan y = \sec x + \tan x \] Now, differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left( \tan y \right) = \frac{d}{dx} \left( \sec x + \tan x \right) \] Using the chain rule on the left-hand side: \[ \sec^2 y \cdot \frac{dy}{dx} = \sec x \tan x + \sec^2 x \] We know from the identity \( \sec^2 y = 1 + \tan^2 y \), and since \( \tan y = \sec x + \tan x \), we substitute: \[ \sec^2 y \cdot \frac{dy}{dx} = \sec^2 x \] Thus: \[ \frac{dy}{dx} = \frac{\sec^2 x}{\sec^2 y} \] Since \( \tan y = \sec x + \tan x \), this simplifies to \( \frac{1}{2} \). Hence, the derivative of \( \tan^{-1} (\sec x + \tan x) \) with respect to \( x \) is equal to \( \frac{1}{2} \).