Find an equation of the tangent line to the graph of the function at the given point.
y = arcsec 14x,

2
14
,
𝜋
4

y =

We are tasked with finding the equation of the tangent line to the function \( y = \arcsec(14x) \) at the point \( \left( \frac{2}{14}, \frac{\pi}{4} \right) \).

### Steps:

1. **Differentiate the function to find the slope**:
   To find the slope of the tangent line, we need to compute the derivative of the function \( y = \arcsec(14x) \).

   The derivative of \( y = \arcsec(u) \) is given by:
   \[
   \frac{d}{dx} \left( \arcsec(u) \right) = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx}
   \]
   Here, \( u = 14x \), so:
   \[
   \frac{dy}{dx} = \frac{1}{|14x| \sqrt{(14x)^2 - 1}} \cdot 14
   \]

   Simplifying:
   \[
   \frac{dy}{dx} = \frac{14}{|14x| \sqrt{196x^2 - 1}}
   \]

2. **Evaluate the slope at the given point**:
   The point of interest is \( \left( \frac{2}{14}, \frac{\pi}{4} \right) \), where \( x = \frac{2}{14} = \frac{1}{7} \).

   Substituting \( x = \frac{1}{7} \) into the derivative:
   \[
   \frac{dy}{dx} \bigg|_{x = \frac{1}{7}} = \frac{14}{|14 \cdot \frac{1}{7}| \sqrt{196 \left( \frac{1}{7} \right)^2 - 1}}
   = \frac{14}{2 \sqrt{4 - 1}} = \frac{14}{2 \cdot \sqrt{3}} = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3}
   \]

3. **Point-slope form of the tangent line**:
   Now that we have the slope \( m = \frac{7\sqrt{3}}{3} \), we use the point-slope form of a line:
   \[
   y - y_1 = m(x - x_1)
   \]
   Substituting \( \left( x_1, y_1 \right) = \left( \frac{2}{14}, \frac{\pi}{4} \right) \) and \( m = \frac{7\sqrt{3}}{3} \):
   \[
   y - \frac{\pi}{4} = \frac{7\sqrt{3}}{3} \left( x - \frac{1}{7} \right)
   \]

This is the equation of the tangent line to the curve at the given point.

Would you like more details or clarifications on this?

### 5 Related Questions:
1. How do we derive the derivative of \( \arcsec(u) \)?
2. What is the geometric interpretation of the arcsecant function?
3. How does the chain rule apply in differentiating composite functions like \( \arcsec(14x) \)?
4. How would the tangent line change for different points on the curve?
5. What are the applications of tangent lines in calculus?

### Tip:
Always verify the points and slopes carefully when using the point-slope form to ensure precision in your equation.

Find an equation of the tangent line to the graph of the function at the given point: y = arcsec(14x), at the point (2/14, π/4).

This solution explains how to find the tangent line to the function y = arcsec(14x) at the point (2/14, π/4). The process includes calculating the derivative using the chain rule and inverse trigonometric functions, followed by applying the point-slope formula. Ideal for students in college-level calculus.

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