Question

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

• A. \(x+\frac{a}{x}\)
• B. \(\frac{x^2+a^2}{ax}\)
• C. \(\frac{x}{a}-\frac{a}{x}\)
• D. \(\frac{a}{x}-\frac{x}{a}\)

Hint:

When expressions contain square-root pairs like \( \sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}} \), differentiation often simplifies using the identity \((p+q)(p-q)=p^2-q^2\).

Answer 1

The Correct Option is C

Step 1: Write the given function.
\[ y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}. \]

Step 2: Express in powers.
\[ y=\frac{x^{1/2}}{a^{1/2}}+a^{1/2}x^{-1/2}. \]

Step 3: Differentiate with respect to \(x\).
\[ \frac{dy}{dx}=\frac{1}{2a^{1/2}}x^{-1/2}-\frac{a^{1/2}}{2}x^{-3/2}. \]

Step 4: Multiply by \(2x\).
\[ 2x\frac{dy}{dx} = 2x\left(\frac{1}{2a^{1/2}}x^{-1/2}-\frac{a^{1/2}}{2}x^{-3/2}\right). \]
\[ 2x\frac{dy}{dx} = \sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}. \]

Step 5: Multiply by \(y\).
Now,
\[ 2xy\frac{dy}{dx} = \left(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\right) \left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right). \]

Step 6: Use identity \((p+q)(p-q)=p^2-q^2\).
\[ 2xy\frac{dy}{dx} = \left(\sqrt{\frac{x}{a}}\right)^2 - \left(\sqrt{\frac{a}{x}}\right)^2. \]
\[ 2xy\frac{dy}{dx} = \frac{x}{a}-\frac{a}{x}. \]

Step 7: Final conclusion.
Thus, the required value is:
\[ \frac{x}{a}-\frac{a}{x}. \]
Final Answer:
\[ \boxed{\frac{x}{a}-\frac{a}{x}}. \]