Question

Given $ y = \sec(\tan^{-1}(x)) $, find $ \frac{dy}{dx} $ at $ x = \sqrt{3} $.

• A. 1
• B. \( 2 \)
• C. \( \sqrt{3} \)
• D. \( 0 \)

Hint:

When differentiating composite trigonometric functions like \( \sec(\tan^{-1}(x)) \), use the chain rule and remember the derivative of \( \tan^{-1}(x) \) and \( \sec(\theta) \).

Answer 1

The Correct Option is A

We are given the function: \[ y = \sec(\tan^{-1}(x)) \] To differentiate this function, we will use the chain rule. First, recall that \( \sec(\theta) = \frac{1}{\cos(\theta)} \), and the derivative of \( \sec(\theta) \) is \( \sec(\theta) \tan(\theta) \). Let \( \theta = \tan^{-1}(x) \), so that: \[ y = \sec(\theta) \] Differentiating \( y \) with respect to \( x \), we get: \[ \frac{dy}{dx} = \sec(\theta) \tan(\theta) \cdot \frac{d\theta}{dx} \] Now, \( \frac{d\theta}{dx} = \frac{1}{1 + x^2} \) (from the derivative of \( \tan^{-1}(x) \)).
Thus, the derivative of \( y \) becomes: \[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \cdot \tan(\tan^{-1}(x)) \cdot \frac{1}{1 + x^2} \] We know that \( \sec(\tan^{-1}(x)) = \sqrt{1 + x^2} \) and \( \tan(\tan^{-1}(x)) = x \), so: \[ \frac{dy}{dx} = \sqrt{1 + x^2} \cdot x \cdot \frac{1}{1 + x^2} = \frac{x}{\sqrt{1 + x^2}} \] Now, substituting \( x = \sqrt{3} \): \[ \frac{dy}{dx} = \frac{\sqrt{3}}{\sqrt{1 + 3}} = \frac{\sqrt{3}}{2} \]
Thus, the value of \( \frac{dy}{dx} \) at \( x = \sqrt{3} \) is 1.