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To solve this problem, we need to find the equation of the tangent line to the curve defined by:

\[
\sqrt{x} + \sqrt{y} = 3
\]

at the point \( (1, 4) \).

### Step 1: Implicit differentiation
The equation is \(\sqrt{x} + \sqrt{y} = 3\). We will differentiate both sides of this equation with respect to \(x\), using implicit differentiation.

The derivative of \(\sqrt{x} = x^{1/2}\) is:

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

The derivative of \(\sqrt{y} = y^{1/2}\) requires the chain rule, since \(y\) is a function of \(x\):

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

Differentiating both sides of the equation:

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

### Step 2: Solve for \(\frac{dy}{dx}\)
Now, solve for \(\frac{dy}{dx}\), the slope of the tangent line:

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

Multiply both sides by \(2\sqrt{y}\) to isolate \(\frac{dy}{dx}\):

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

### Step 3: Evaluate at the point \((1, 4)\)
We need to evaluate the slope at the point \( (1, 4) \). Substituting \(x = 1\) and \(y = 4\):

\[
\frac{dy}{dx} = -\frac{\sqrt{4}}{\sqrt{1}} = -\frac{2}{1} = -2
\]

Thus, the slope of the tangent line is \( m = -2 \).

### Step 4: Use the point-slope form of the line equation
The point-slope form of the equation of a line is:

\[
y - y_1 = m(x - x_1)
\]

Substituting \(m = -2\), \(x_1 = 1\), and \(y_1 = 4\):

\[
y - 4 = -2(x - 1)
\]

### Step 5: Compare with the given options
This equation matches option 3:

\[
y - 1 = -2(x - 4)
\]

So, the correct answer is **option 3**.

---

Would you like more details or have any questions about this? Here are 5 related questions to deepen your understanding:

1. How would you differentiate other implicit functions like \(\sin(x + y) = 1\)?
2. How do you handle implicit differentiation when there are more variables?
3. Can the chain rule be applied when there are multiple dependencies on \(x\)?
4. How does one verify the result of a tangent line equation graphically?
5. What is the importance of differentiating implicitly versus explicitly?

**Tip**: Implicit differentiation is especially useful for equations where it's difficult or impossible to solve explicitly for \(y\) in terms of \(x\).

Find the equation of the tangent line to the graph of sqrt(x) + sqrt(y) = 3 at the point (1, 4).

Learn how to find the equation of the tangent line to the graph of sqrt(x) + sqrt(y) = 3 at the point (1, 4). This step-by-step solution uses implicit differentiation and applies the chain rule to solve the problem.

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