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To find the equation of the tangent line to the function \( f(x) = \frac{\sqrt{x}}{7x - 1} \) at the point \( (1, f(1)) \), we need to follow these steps:

1. **Evaluate \( f(1) \)** to get the \( y \)-coordinate of the point on the curve.
2. **Compute \( f'(x) \)**, the derivative of \( f(x) \), to find the slope of the tangent line at \( x = 1 \).
3. Use the point-slope form of a line equation \( y - y_1 = m(x - x_1) \) with \( (x_1, y_1) = (1, f(1)) \) and \( m = f'(1) \).

### Step 1: Evaluate \( f(1) \)
\[
f(1) = \frac{\sqrt{1}}{7 \cdot 1 - 1} = \frac{1}{7 - 1} = \frac{1}{6}.
\]
So, the point is \( (1, \frac{1}{6}) \).

### Step 2: Compute \( f'(x) \)
To find \( f'(x) \), use the quotient rule. Let:
\[
u = \sqrt{x} = x^{1/2} \quad \text{and} \quad v = 7x - 1.
\]
Then,
\[
u' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \quad \text{and} \quad v' = 7.
\]

Using the quotient rule \( f'(x) = \frac{u'v - uv'}{v^2} \),
\[
f'(x) = \frac{\left(\frac{1}{2\sqrt{x}}\right)(7x - 1) - (\sqrt{x})(7)}{(7x - 1)^2}.
\]
Simplifying each term:
\[
f'(x) = \frac{\frac{7x - 1}{2\sqrt{x}} - 7\sqrt{x}}{(7x - 1)^2}.
\]
Rewrite \( 7\sqrt{x} \) as \( \frac{7x}{\sqrt{x}} \):
\[
f'(x) = \frac{\frac{7x - 1 - 14x}{2\sqrt{x}}}{(7x - 1)^2} = \frac{\frac{-13x - 1}{2\sqrt{x}}}{(7x - 1)^2}.
\]

Now, substitute \( x = 1 \) to find \( f'(1) \):
\[
f'(1) = \frac{\frac{-13 \cdot 1 - 1}{2 \cdot 1}}{(7 \cdot 1 - 1)^2} = \frac{\frac{-14}{2}}{6^2} = \frac{-7}{36}.
\]

### Step 3: Equation of the Tangent Line
With \( m = f'(1) = -\frac{7}{36} \) and point \( (1, \frac{1}{6}) \), we use the point-slope form:
\[
y - \frac{1}{6} = -\frac{7}{36}(x - 1).
\]
Expanding and simplifying:
\[
y = -\frac{7}{36}x + \frac{7}{36} + \frac{1}{6}.
\]
Convert \( \frac{1}{6} \) to \( \frac{6}{36} \):
\[
y = -\frac{7}{36}x + \frac{7 + 6}{36} = -\frac{7}{36}x + \frac{13}{36}.
\]

### Final Answer
\[
y = -\frac{7}{36}x + \frac{13}{36}.
\]

Would you like further details on any step, or have any questions?

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Here are five related questions you might explore:
1. How would the process change if the function was more complex, like a product instead of a quotient?
2. What is the significance of the tangent line in terms of the behavior of \( f(x) \) near \( x = 1 \)?
3. Can we find a general formula for the slope of the tangent for functions of the form \( \frac{\sqrt{x}}{ax - b} \)?
4. How does the value of the derivative \( f'(1) \) affect the steepness of the tangent line?
5. What is the geometric interpretation of the sign of the derivative?

**Tip:** When using the quotient rule, make sure to carefully apply and simplify each term, especially when involving square roots.

Find an equation for the line tangent to the graph of f(x) = sqrt(x) / (7x - 1) at the point (1, f(1)).

Find the tangent line to the function f(x) = sqrt(x) / (7x - 1) at the point (1, f(1)). This solution involves using the derivative and quotient rule to find the slope and applying the point-slope form for the line equation. Ideal for advanced calculus students in Grades 11-12.

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