If `sqrt(x)+sqrt(y) = sqrt(a)`, find `(d^(2)y)/(dx^(2))` at x = a.

Answer

To solve the problem, we start with the equation given:

\[ \sqrt{x} + \sqrt{y} = \sqrt{a} \]

We need to find the second derivative of \( y \) with respect to \( x \), denoted as \( \frac{d^2y}{dx^2} \), at the point where \( x = a \).

### Step 1: Differentiate the equation with respect to \( x \)

Differentiating both sides with respect to \( x \):

\[ \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(\sqrt{a}) \]

Since \( a \) is a constant, its derivative is 0. The derivatives of \( \sqrt{x} \) and \( \sqrt{y} \) are:

\[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \]

### Step 2: Solve for \( \frac{dy}{dx} \)

Rearranging the equation gives:

\[ \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \]

Multiplying both sides by \( 2\sqrt{y} \):

\[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \]

### Step 3: Differentiate again to find \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \) with respect to \( x \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{\sqrt{y}}{\sqrt{x}}\right) \]

Using the quotient rule, where \( u = \sqrt{y} \) and \( v = \sqrt{x} \):

\[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

\[ \frac{du}{dx} = \frac{1}{2\sqrt{y}} \frac{dy}{dx}, \quad \frac{dv}{dx} = \frac{1}{2\sqrt{x}} \]

Substituting these into the quotient rule gives:

\[ \frac{d^2y}{dx^2} = \frac{\sqrt{x} \left(\frac{1}{2\sqrt{y}} \frac{dy}{dx}\right) - \sqrt{y} \left(\frac{1}{2\sqrt{x}}\right)}{x} \]

### Step 4: Substitute \( \frac{dy}{dx} \) into the equation

Substituting \( \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \):

\[ \frac{d^2y}{dx^2} = \frac{\sqrt{x} \left(\frac{1}{2\sqrt{y}} \left(-\frac{\sqrt{y}}{\sqrt{x}}\right)\right) - \sqrt{y} \left(\frac{1}{2\sqrt{x}}\right)}{x} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = \frac{-\frac{1}{2\sqrt{x}} - \frac{\sqrt{y}}{2\sqrt{x}}}{x} \]

### Step 5: Evaluate at \( x = a \)

Now we substitute \( x = a \):

\[ \frac{d^2y}{dx^2} = \frac{-\frac{1}{2\sqrt{a}} - \frac{\sqrt{y}}{2\sqrt{a}}}{a} \]

Since \( \sqrt{y} = \sqrt{a} - \sqrt{a} = 0 \):

\[ \frac{d^2y}{dx^2} = \frac{-\frac{1}{2\sqrt{a}}}{a} \]

Thus:

\[ \frac{d^2y}{dx^2} = \frac{-1}{2a\sqrt{a}} \]

### Final Result

At \( x = a \):

\[ \frac{d^2y}{dx^2} = \frac{1}{2a} \]