If `sqrt(x)+sqrt(y)=2` then what is `(dy)/(dx)` at `y=1` equal to ?

Options

A. 5
B. 4
C. 2
D. `-1`

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given the equation \(\sqrt{x} + \sqrt{y} = 2\) at the point where \(y = 1\).

### Step-by-Step Solution:

1. **Differentiate the equation**:
Start with the equation:
\[ \sqrt{x} + \sqrt{y} = 2 \]
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(2) \]
This gives:
\[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \]

2. **Rearrange the equation**:
Isolate \(\frac{dy}{dx}\):
\[ \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \]
Multiply both sides by \(2\sqrt{y}\):
\[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \]

3. **Find \(\sqrt{x}\) when \(y = 1\)**:
Substitute \(y = 1\) into the original equation:
\[ \sqrt{x} + \sqrt{1} = 2 \]
This simplifies to:
\[ \sqrt{x} + 1 = 2 \]
Therefore:
\[ \sqrt{x} = 1 \quad \Rightarrow \quad x = 1 \]

4. **Substitute \(y = 1\) and \(x = 1\) into \(\frac{dy}{dx}\)**:
Now substitute \(y = 1\) and \(x = 1\) into the equation for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{\sqrt{1}}{\sqrt{1}} = -\frac{1}{1} = -1 \]

### Final Answer:
Thus, \(\frac{dy}{dx}\) at \(y = 1\) is:
\[ \frac{dy}{dx} = -1 \]