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.