if ` sqrt(x+y)+sqrt(y-x)=c ` then ` (d^2 y)/(dx^2)` equals

Options

A. `2//c`
B. `-2//c^2`
C. `2//c^2`
D. None of these

Answer

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

\[ \sqrt{x+y} + \sqrt{y-x} = c \]

### Step 1: Rationalize the equation
We can rewrite the equation by isolating one of the square roots. Let's isolate \(\sqrt{y-x}\):

\[ \sqrt{y-x} = c - \sqrt{x+y} \]

### Step 2: Square both sides
Now, we square both sides to eliminate the square root:

\[ y - x = (c - \sqrt{x+y})^2 \]

Expanding the right-hand side:

\[ y - x = c^2 - 2c\sqrt{x+y} + (x+y) \]

### Step 3: Rearranging the equation
Rearranging gives us:

\[ y - x - (x+y) = c^2 - 2c\sqrt{x+y} \]

This simplifies to:

\[ -y = c^2 - 2c\sqrt{x+y} \]

### Step 4: Isolate the square root
Now, isolate the square root:

\[ 2c\sqrt{x+y} = c^2 + y \]

### Step 5: Square again
Square both sides again:

\[ (2c\sqrt{x+y})^2 = (c^2 + y)^2 \]

This expands to:

\[ 4c^2(x+y) = c^4 + 2c^2y + y^2 \]

### Step 6: Rearranging the equation
Rearranging gives us:

\[ 4c^2x + 4c^2y = c^4 + 2c^2y + y^2 \]

### Step 7: Collect like terms
Collecting like terms results in:

\[ 4c^2x + 4c^2y - 2c^2y - y^2 = c^4 \]

This simplifies to:

\[ 4c^2x + 2c^2y - y^2 = c^4 \]

### Step 8: Differentiate with respect to \(x\)
Now we differentiate both sides with respect to \(x\):

\[ \frac{d}{dx}(4c^2x + 2c^2y - y^2) = \frac{d}{dx}(c^4) \]

The right-hand side is zero since \(c\) is a constant. The left-hand side becomes:

\[ 4c^2 + 2c^2\frac{dy}{dx} - 2y\frac{dy}{dx} = 0 \]

### Step 9: Solve for \(\frac{dy}{dx}\)
Rearranging gives:

\[ 2c^2\frac{dy}{dx} - 2y\frac{dy}{dx} = -4c^2 \]

Factoring out \(\frac{dy}{dx}\):

\[ \frac{dy}{dx}(2c^2 - 2y) = -4c^2 \]

So,

\[ \frac{dy}{dx} = \frac{-4c^2}{2(c^2 - y)} = \frac{-2c^2}{c^2 - y} \]

### Step 10: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Now we differentiate \(\frac{dy}{dx}\) again with respect to \(x\):

Using the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{(c^2 - y)(-2\frac{dy}{dx}) - (-2c^2)(\frac{dy}{dx})}{(c^2 - y)^2} \]

Substituting \(\frac{dy}{dx}\) into the equation will yield the final result.

### Final Result
After simplification, we find:

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