Question

If \( y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right) \), then \( y'(1) \) is equal to:

• A. 0
• B. \( \frac{1}{2} \)
• C. -1
• D. \(-\frac{1}{4}\)

Hint:

To differentiate inverse trigonometric functions, remember the chain rule and handle the rational expressions carefully.

Answer 1

The Correct Option is D

Step 1: Rewrite the given equation using the inverse tangent identity
We are given \( y = \tan^{-1}\left(\frac{\sqrt{x} - x}{1 + \sqrt{x} \cdot x}\right) \). Recall the identity: \[ \tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1}a - \tan^{-1}b \] Using this identity, we can rewrite the given equation as: \[ y = \tan^{-1}(\sqrt{x}) - \tan^{-1}(x) \] Step 2: Differentiate \( y \) with respect to \( x \)
Differentiate both sides with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}(\sqrt{x}) - \tan^{-1}(x) \right) \] \[ y'(x) = \frac{d}{dx} \tan^{-1}(\sqrt{x}) - \frac{d}{dx} \tan^{-1}(x) \] Recall that \( \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \frac{du}{dx} \). \[ y'(x) = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x}) - \frac{1}{1 + x^2} \] \[ y'(x) = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} - \frac{1}{1 + x^2} \] Step 3: Evaluate \( y'(1) \)
Substitute \( x = 1 \) into the expression for \( y'(x) \): \[ y'(1) = \frac{1}{1 + 1} \cdot \frac{1}{2\sqrt{1}} - \frac{1}{1 + 1^2} \] \[ y'(1) = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \] \[ y'(1) = \frac{1}{4} - \frac{1}{2} \] \[ y'(1) = \frac{1}{4} - \frac{2}{4} \] \[ y'(1) = -\frac{1}{4} \] Therefore, \( y'(1) = -\frac{1}{4} \).