Question

If \( y = \sec(\tan^{-1} x) \), find \( \frac{dy}{dx} \), given that \( x = 1 \):

• A. 2
• B. 1
• C. 0
• D. Undefined

Hint:

To differentiate expressions involving inverse trigonometric functions, first express the function in terms of simpler trigonometric identities, then apply the chain rule.

Answer 1

The Correct Option is A

We are given that \( y = \sec(\tan^{-1} x) \). To find \( \frac{dy}{dx} \), we will differentiate implicitly with respect to \( x \). Step 1: Use the chain rule. We know that: \[ y = \sec(\tan^{-1} x). \] Let \( \theta = \tan^{-1} x \), so that \( y = \sec(\theta) \). Now differentiate \( y = \sec(\theta) \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \sec(\theta) = \sec(\theta) \tan(\theta) \frac{d\theta}{dx}. \] Next, we need to find \( \frac{d\theta}{dx} \).
Step 2: Differentiate \( \theta = \tan^{-1} x \). We know that: \[ \frac{d}{dx} \left( \tan^{-1} x \right) = \frac{1}{1 + x^2}. \] Thus, \[ \frac{d\theta}{dx} = \frac{1}{1 + x^2}. \]
Step 3: Substitute back to find \( \frac{dy}{dx} \). Substituting \( \frac{d\theta}{dx} = \frac{1}{1 + x^2} \) into the expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \sec(\theta) \tan(\theta) \cdot \frac{1}{1 + x^2}. \] Now we need to express \( \sec(\theta) \) and \( \tan(\theta) \) in terms of \( x \).
Step 4: Find expressions for \( \sec(\theta) \) and \( \tan(\theta) \). Since \( \theta = \tan^{-1} x \), we know that: \[ \tan(\theta) = x. \] Using the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we find: \[ \sec(\theta) = \sqrt{1 + x^2}. \] Thus, the derivative becomes: \[ \frac{dy}{dx} = \sqrt{1 + x^2} \cdot x \cdot \frac{1}{1 + x^2}. \] Simplifying: \[ \frac{dy}{dx} = \frac{x}{\sqrt{1 + x^2}}. \]
Step 5: Evaluate at \( x = 1 \). Now, substitute \( x = 1 \) into the derivative: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}. \] So the correct answer is \( \boxed{2} \).