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, \( y'(1) = -\frac{1}{4} \).

Answer 2

Step 1: Differentiate using the chain rule
We are given \( y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right) \). To find \( y'(1) \), we differentiate the function using the chain rule.

The derivative of \( \tan^{-1}(u) \) is: \[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] Let \( u = \frac{\sqrt{x} - x}{1 + x^{3/2}} \). Then: \[ y' = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] We need to find \( \frac{du}{dx} \).

Step 2: Differentiate \( u = \frac{\sqrt{x} - x}{1 + x^{3/2}} \)
We apply the quotient rule to differentiate \( u \). The quotient rule is: \[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2} \] where \( f(x) = \sqrt{x} - x \) and \( g(x) = 1 + x^{3/2} \). First, we differentiate \( f(x) = \sqrt{x} - x \): \[ f'(x) = \frac{1}{2\sqrt{x}} - 1 \] Next, we differentiate \( g(x) = 1 + x^{3/2} \): \[ g'(x) = \frac{3}{2} x^{1/2} \] Now, applying the quotient rule: \[ \frac{du}{dx} = \frac{(1 + x^{3/2}) \left( \frac{1}{2\sqrt{x}} - 1 \right) - (\sqrt{x} - x) \left( \frac{3}{2} x^{1/2} \right)}{(1 + x^{3/2})^2} \] Step 3: Simplify and evaluate at \( x = 1 \)
Now, we evaluate \( y' \) at \( x = 1 \). First, compute \( u \) at \( x = 1 \): \[ u = \frac{\sqrt{1} - 1}{1 + 1^{3/2}} = \frac{1 - 1}{1 + 1} = 0 \] So, at \( x = 1 \), \( u = 0 \). This simplifies \( y' \) to: \[ y' = \frac{1}{1 + 0^2} \cdot \frac{du}{dx} = \frac{du}{dx} \] Now, substitute \( x = 1 \) into the expression for \( \frac{du}{dx} \): \[ \frac{du}{dx} = \frac{(1 + 1) \left( \frac{1}{2\sqrt{1}} - 1 \right) - (1 - 1) \left( \frac{3}{2} \cdot 1^{1/2} \right)}{(1 + 1)^2} \] Simplifying: \[ \frac{du}{dx} = \frac{2 \left( \frac{1}{2} - 1 \right)}{4} = \frac{2 \cdot (-\frac{1}{2})}{4} = -\frac{1}{4} \] Therefore, the value of \( y'(1) \) is:

\(-\frac{1}{4}\)