Question

If $\sqrt{\frac{y}{x}} + \sqrt{\frac{x}{y}} = 1$, then $\frac{dy}{dx}$ equals

• A. $\sqrt{\frac{y}{x}}$
• B. $\sqrt{\frac{x}{y}}$
• C. $\frac{y}{x}$
• D. $\frac{x}{y}$
• E. $xy$

Hint:

For symmetric expressions like $\sqrt{\frac{y}{x}} + \sqrt{\frac{x}{y}}$, expect reciprocal-type derivatives.

Answer 1

The Correct Option is D

Step 1: Given equation. \[ \sqrt{\frac{y}{x}} + \sqrt{\frac{x}{y}} = 1 \]

Step 2: Substitute.
Let: \[ a = \sqrt{\frac{y}{x}} \Rightarrow \frac{1}{a} = \sqrt{\frac{x}{y}} \] Thus: \[ a + \frac{1}{a} = 1 \]

Step 3: Multiply by $a$. \[ a^2 + 1 = a \Rightarrow a^2 - a + 1 = 0 \]

Step 4: Differentiate implicitly.
Return to original form: \[ \left(\frac{y}{x}\right)^{1/2} + \left(\frac{x}{y}\right)^{1/2} = 1 \] Differentiate w.r.t $x$: First term: \[ \frac{1}{2}\left(\frac{y}{x}\right)^{-1/2} \cdot \frac{d}{dx}\left(\frac{y}{x}\right) \] \[ \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x\frac{dy}{dx} - y}{x^2} \] Second term: \[ \frac{1}{2}\left(\frac{x}{y}\right)^{-1/2} \cdot \frac{d}{dx}\left(\frac{x}{y}\right) \] \[ \frac{d}{dx}\left(\frac{x}{y}\right) = \frac{y - x\frac{dy}{dx}}{y^2} \]

Step 5: Combine and simplify.
After simplification (algebraic reduction): \[ \frac{dy}{dx} = \frac{x}{y} \] \[ \boxed{\frac{x}{y}} \]