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The problem from your screenshot is asking for the equation of the tangent line to the function \( f(x) = 4 \tan(x) \) at \( x = \frac{\pi}{4} \).

### Steps to solve:
1. **Find the derivative** of \( f(x) = 4 \tan(x) \), which gives us the slope of the tangent line at any point \( x \).

\[
f'(x) = 4 \sec^2(x)
\]

2. **Evaluate the derivative at \( x = \frac{\pi}{4} \)** to find the slope of the tangent line at this specific point.

\[
f'\left(\frac{\pi}{4}\right) = 4 \sec^2\left(\frac{\pi}{4}\right)
\]

Since \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \), we have:

\[
f'\left(\frac{\pi}{4}\right) = 4 \times \left(\sqrt{2}\right)^2 = 4 \times 2 = 8
\]

So, the slope of the tangent line is 8.

3. **Find the point on the curve** at \( x = \frac{\pi}{4} \):

\[
f\left(\frac{\pi}{4}\right) = 4 \tan\left(\frac{\pi}{4}\right) = 4 \times 1 = 4
\]

So, the point is \( \left(\frac{\pi}{4}, 4\right) \).

4. **Use the point-slope form** of a line:

The equation of the tangent line is given by:

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

Where \( m = 8 \), \( (x_1, y_1) = \left(\frac{\pi}{4}, 4\right) \).

\[
y - 4 = 8 \left( x - \frac{\pi}{4} \right)
\]

Simplifying:

\[
y = 8x - 2\pi + 4
\]

So, the equation of the tangent line is:

\[
y = 8x - 2\pi + 4
\]

Would you like further details or clarifications on this? 

Here are some related questions for further practice:
1. What is the equation of the tangent line to \( f(x) = \sin(x) \) at \( x = \frac{\pi}{6} \)?
2. How do you find the derivative of \( f(x) = \tan(x) \)?
3. What is the significance of the tangent line in the context of a graph of a function?
4. How would the tangent line change if \( f(x) = 5\tan(x) \) instead?
5. Can you describe what happens to the tangent line as \( x \) approaches \( \pi \)?

**Tip**: Always remember that the tangent line at a given point has the same slope as the derivative of the function at that point.

Given f(x) = 4tan(x), write the equation of the line tangent to y = f(x) when x = π/4.

Learn how to find the equation of the tangent line to the trigonometric function f(x) = 4tan(x) at the point where x = π/4. This solution involves calculating the derivative, evaluating at the given point, and applying the point-slope form. Suitable for high school calculus students.

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