Question

If \[ \sqrt{x}+\sqrt{y}=\sqrt{a}, \] then \[ \left(\frac{d^2y}{dx^2}\right)_{x=a} = \]

• A. \(\dfrac{1}{a}\)
• B. \(\dfrac{1}{2a}\)
• C. \(\dfrac{1}{2\sqrt a}\)
• D. \(\dfrac{1}{2}\)

Hint:

When an equation involves square roots of both variables, isolate one root and square the equation to obtain an explicit relation before differentiating.

Answer 1

The Correct Option is B

Concept: First express \(y\) explicitly in terms of \(x\), then differentiate twice and substitute \(x=a\).

Step 1: Express \(y\) as a function of \(x\). Given, \[ \sqrt{x}+\sqrt{y}=\sqrt{a}. \] Therefore, \[ \sqrt{y} = \sqrt{a}-\sqrt{x}. \] Squaring both sides, \[ y = (\sqrt{a}-\sqrt{x})^2. \] \[ y = a+x-2\sqrt{ax}. \]

Step 2: Find the first derivative. Differentiating with respect to \(x\), \[ \frac{dy}{dx} = 1-2\sqrt{a}\frac{d}{dx}\left(x^{1/2}\right). \] \[ = 1-\frac{\sqrt a}{\sqrt x}. \]

Step 3: Find the second derivative. Differentiating again, \[ \frac{d^2y}{dx^2} = -\sqrt a\frac{d}{dx}\left(x^{-1/2}\right). \] \[ = -\sqrt a\left(-\frac12x^{-3/2}\right). \] \[ = \frac{\sqrt a}{2x^{3/2}}. \]

Step 4: Evaluate at \(x=a\). \[ \left(\frac{d^2y}{dx^2}\right)_{x=a} = \frac{\sqrt a}{2a^{3/2}}. \] \[ = \frac{1}{2a}. \] Therefore, \[ \boxed{\left(\frac{d^2y}{dx^2}\right)_{x=a}=\frac{1}{2a}} \] \[ \boxed{\text{Answer = (B)}} \]