Question

If \(\sqrt{y+x}+\sqrt{y-x} = c\) such that \(\frac{dy}{dx} = f(x)-\sqrt{[f(x)]^2-1}\), then \(f(x) =\) ____

• A. \(\frac{y}{x}\)
• B. \(-\frac{x}{y}\)
• C. \(-\frac{y}{x}\)
• D. \(\frac{x}{y}\)

Hint:

Differentiate implicitly and rationalise to get \(\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}\).

Answer 1

The Correct Option is A

Step 1: Differentiate:
From \(\sqrt{y+x}+\sqrt{y-x} = c\): \(\dfrac{y'+1}{2\sqrt{y+x}} + \dfrac{y'-1}{2\sqrt{y-x}} = 0\).
Cross multiply: \((y'+1)\sqrt{y-x} + (y'-1)\sqrt{y+x} = 0\).

Step 2: Solve for y':
\[ y'\left(\sqrt{y-x}+\sqrt{y+x}\right) = \sqrt{y+x} - \sqrt{y-x} \]
\[ y' = \frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}} \]

Step 3: Rationalise:
Multiply top and bottom by \(\sqrt{y+x}-\sqrt{y-x}\). The denominator becomes \((y+x)-(y-x) = 2x\).
The numerator becomes \((y+x) + (y-x) - 2\sqrt{y^2-x^2} = 2y - 2\sqrt{y^2-x^2}\).
\[ y' = \frac{y - \sqrt{y^2-x^2}}{x} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2}-1} \]

Step 4: Compare:
We are told \(\frac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}\). Matching gives \(f(x) = \frac{y}{x}\).

Final Answer:
\(f(x) = \frac{y}{x}\), option (A). \[ \boxed{\frac{y}{x}} \]