If `sqrt(x)+sqrt(y)=sqrt(a)," then "y_(2)" at "x=a` is

Options

A. a
B. `(1)/(2a)`
C. `(1)/(a)`
D. `sqrt(a)`

Answer

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

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

We need to find the second derivative \(y''\) at \(x = a\).

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

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

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

### Step 2: Rearranging the equation

From the above equation, we can isolate \(\frac{dy}{dx}\):

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

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

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

### Step 3: Find the second derivative \(y''\)

Now we need to differentiate \(\frac{dy}{dx}\) to find \(y''\):

\[ y' = -\frac{\sqrt{y}}{\sqrt{x}} \]

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

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

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

- \(\frac{du}{dx} = -\frac{1}{2\sqrt{y}} \frac{dy}{dx}\)
- \(\frac{dv}{dx} = \frac{1}{2\sqrt{x}}\)

Substituting these into the quotient rule:

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

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

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

\[ y'' = \frac{\sqrt{x} \left(-\frac{1}{2\sqrt{y}} \left(-\frac{\sqrt{y}}{\sqrt{x}}\right)\right) + \frac{\sqrt{y}}{2\sqrt{x}}}{x} \]

This simplifies to:

\[ y'' = \frac{\sqrt{x} \cdot \frac{1}{2\sqrt{y}} \cdot \frac{\sqrt{y}}{\sqrt{x}} + \frac{\sqrt{y}}{2\sqrt{x}}}{x} \]

### Step 5: Simplifying the expression

The terms simplify:

\[ y'' = \frac{1}{2x} + \frac{\sqrt{y}}{2x\sqrt{x}} \]

### Step 6: Substitute \(x = a\)

Now we substitute \(x = a\):

From the original equation, we have:

\[ \sqrt{y} = \sqrt{a} - \sqrt{a} = 0 \]

Thus, we can substitute back into our expression for \(y''\):

\[ y'' = \frac{1}{2a} + \frac{0}{2a\sqrt{a}} = \frac{1}{2a} \]

### Final Answer

Thus, the second derivative \(y''\) at \(x = a\) is:

\[ \boxed{\frac{1}{2a}} \]