Question

If \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \), then \( 2x \frac{dy}{dx} \) is equal to

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

Hint:

When differentiating expressions with fractions, converting everything to a standard \( x^n \) format first makes the process much more straightforward.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the derivative of the given function \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \), we can rewrite the terms using exponents: \( y = x^{1/2} + x^{-1/2} \). We will then use the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) to differentiate with respect to \( x \).

Step 2: Detailed Explanation:
Given \( y = x^{1/2} + x^{-1/2} \).
Differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2}x^{1/2 - 1} - \frac{1}{2}x^{-1/2 - 1} \] \[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \] \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} \]

Step 3: Further Calculation:
Now, we multiply the derivative by \( 2x \): \[ 2x \frac{dy}{dx} = 2x \left( \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} \right) \] \[ 2x \frac{dy}{dx} = \frac{2x}{2\sqrt{x}} - \frac{2x}{2x\sqrt{x}} \] \[ 2x \frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}} \]

Step 4: Final Answer:
Thus, \( 2x \frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}} \). This corresponds to option (A).