We are given the function $f(x) = \tan^2(x)$ and the point $\left( \frac{\pi}{4}, 1 \right)$. The task is to find the equation of the tangent line to the graph of this function at that point.

Steps:

1. Find the derivative of $f(x) = \tan^2(x)$: To find the equation of the tangent line, we first need to compute the derivative of the function. We will use the chain rule:

   $f(x) = \tan^2(x)$ The derivative of $f(x)$ is: $f'(x) = 2\tan(x) \cdot \sec^2(x)$

2. Evaluate the derivative at $x = \frac{\pi}{4}$: We substitute $x = \frac{\pi}{4}$ into the derivative: $f'\left( \frac{\pi}{4} \right) = 2 \tan\left( \frac{\pi}{4} \right) \cdot \sec^2\left( \frac{\pi}{4} \right)$ Since $\tan\left( \frac{\pi}{4} \right) = 1$ and $\sec\left( \frac{\pi}{4} \right) = \sqrt{2}$, we get: $f'\left( \frac{\pi}{4} \right) = 2 \times 1 \times 2 = 4$

3. Write the equation of the tangent line: The equation of a tangent line is given by the point-slope form: $y - y_1 = m(x - x_1)$ Here, $m = 4$ (the slope), $x_1 = \frac{\pi}{4}$, and $y_1 = 1$. Substituting these values into the equation: $y - 1 = 4 \left( x - \frac{\pi}{4} \right)$

4. Simplify the equation: $y - 1 = 4x - \pi$ $y = 4x - \pi + 1$ Thus, the equation of the tangent line is: $y = 4x - \pi + 1$

Would you like further details or clarifications on any step?

Related questions:

1. How is the chain rule applied to functions like $\tan^2(x)$?
2. What is the geometric meaning of the tangent line to a curve?
3. How does evaluating a derivative at a point relate to the slope of the tangent line?
4. Can we always use point-slope form to find the equation of a tangent line?
5. What happens if we change the function to something like $\cot^2(x)$?

Tip: The chain rule is essential when differentiating compositions of functions like $\tan^2(x)$, where the outer function is a square and the inner function is $\tan(x)$.