Question

If \[ y = \tan^{-1}(\sec x + \tan x), \quad \text{then} \quad \frac{dy}{dx} = \]

• A. \( \frac{1}{2} \)
• B. 1
• C. \( -\frac{1}{2} \)
• D. -1

Hint:

For functions involving \( \tan^{-1} \) and trigonometric expressions, use the chain rule carefully and simplify the resulting expressions.

Answer 1

The Correct Option is A

Step 1: Differentiating the function.
We are given \( y = \tan^{-1}(\sec x + \tan x) \). To find \( \frac{dy}{dx} \), we use the chain rule. The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \), where \( u = \sec x + \tan x \).

Step 2: Applying the chain rule.
We first differentiate \( u = \sec x + \tan x \). The derivative of \( \sec x \) is \( \sec x \tan x \), and the derivative of \( \tan x \) is \( \sec^2 x \). So, \[ \frac{du}{dx} = \sec x \tan x + \sec^2 x \] Now, applying the chain rule: \[ \frac{dy}{dx} = \frac{1}{1 + (\sec x + \tan x)^2} \cdot (\sec x \tan x + \sec^2 x) \] After simplifying, we find that the correct answer is \( \frac{1}{2} \), corresponding to option (A).

Step 3: Conclusion.
Thus, the correct value of \( \frac{dy}{dx} \) is \( \frac{1}{2} \), making option (A) the correct answer.