Question

Let $y = \tan^{-1}(\sec x + \tan x)$. Then $\frac{dy}{dx} =$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sec x + \tan x}$
• D. $\frac{1}{\sec^2 x}$
• E. $\frac{1}{\tan x}$

Hint:

The transformation $\sec x + \tan x = \tan(\frac{\pi}{4} + \frac{x}{2})$ is a very common identity in calculus. Memorizing it can save several steps of algebraic manipulation.

Answer 1

The Correct Option is B

Concept: Simplify the trigonometric expression inside the inverse tangent function before differentiating. This often reduces a complex chain rule problem to a simple linear derivative.

Step 1: Simplify the inner expression using basic identities.
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \] Using half-angle identities: $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$. \[ \frac{1 + \sin x}{\cos x} = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \]

Step 2: Convert to a tangent form.
Divide numerator and denominator by $\cos \frac{x}{2}$: \[ \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} = \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) \]

Step 3: Substitute back into $y$ and differentiate.
\[ y = \tan^{-1}\left[ \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) \right] = \frac{\pi}{4} + \frac{x}{2} \] Differentiating with respect to $x$: \[ \frac{dy}{dx} = \frac{d}{dx}\left( \frac{\pi}{4} + \frac{x}{2} \right) = 0 + \frac{1}{2} = \frac{1}{2} \]