If `sqrt(x) + sqrt(y) = sqrt(a)`, then `(dy)/(dx) = 1/(2sqrt(x)) + 1/(2sqrt(y)) = 1/(2sqrt(a))`

Answer

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

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

where \(a\) is a constant. We need to differentiate both sides with respect to \(x\) to find \(\frac{dy}{dx}\).

### Step 1: Differentiate both sides
Differentiating both sides with respect to \(x\):

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

### Step 2: Apply the differentiation formula
Using the differentiation formula \(\frac{d}{dx}(x^{n}) = n x^{n-1}\), we differentiate each term:

- For \(\sqrt{x}\), we have:
\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \]

- For \(\sqrt{y}\), since \(y\) is a function of \(x\), we apply the chain rule:
\[ \frac{d}{dx}(\sqrt{y}) = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \]

- For \(\sqrt{a}\), since \(a\) is a constant:
\[ \frac{d}{dx}(\sqrt{a}) = 0 \]

Putting it all together, we have:

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

### Step 3: Solve for \(\frac{dy}{dx}\)
Now, we can isolate \(\frac{dy}{dx}\):

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

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

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

### Step 4: Verify the expression
The original claim was that:

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

From our calculation, we found:

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

This does not match the original claim, indicating that the statement is false.

### Conclusion
Thus, the statement given in the question is **false**.

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