We need to find the equation of the tangent line to the graph of $y = \tan^{-1}\left( \frac{1}{4}x \right)$ at the point $(4, \frac{\pi}{4})$.

Step 1: Recall the point-slope form of a line

The equation of the tangent line can be written in point-slope form: $y - y_1 = m(x - x_1)$ where $(x_1, y_1) = (4, \frac{\pi}{4})$ is the point of tangency, and $m$ is the slope of the tangent line.

Step 2: Find the derivative of the function

To find the slope of the tangent line, we need to differentiate the function $y = \tan^{-1}\left( \frac{1}{4}x \right)$.

Using the chain rule, the derivative of $y = \tan^{-1}(u)$ is: $\frac{d}{dx}\left[ \tan^{-1}(u) \right] = \frac{1}{1 + u^2} \cdot \frac{du}{dx}$ Here, $u = \frac{1}{4}x$, so: $\frac{du}{dx} = \frac{1}{4}$ Thus, the derivative is: $\frac{dy}{dx} = \frac{1}{1 + \left( \frac{1}{4}x \right)^2} \cdot \frac{1}{4}$ Simplifying: $\frac{dy}{dx} = \frac{1}{4} \cdot \frac{1}{1 + \frac{x^2}{16}} = \frac{1}{4 + x^2}$

Step 3: Evaluate the derivative at $x = 4$

To find the slope of the tangent line, substitute $x = 4$ into the derivative: $m = \frac{1}{4 + 4^2} = \frac{1}{4 + 16} = \frac{1}{20}$

Step 4: Write the equation of the tangent line

Now that we know the slope $m = \frac{1}{20}$ and the point of tangency $(4, \frac{\pi}{4})$, we can use the point-slope form: $y - \frac{\pi}{4} = \frac{1}{20}(x - 4)$

Simplifying: $y = \frac{1}{20}x - \frac{4}{20} + \frac{\pi}{4}$ $y = \frac{1}{20}x - \frac{1}{5} + \frac{\pi}{4}$

Thus, the equation of the tangent line is: $y = \frac{1}{20}x + \frac{\pi}{4} - \frac{1}{5}$

Final equation:

$y = \frac{1}{20}x + \frac{\pi}{4} - \frac{1}{5}$

Would you like more details on any part of this?

Here are 5 questions you might explore further:

1. How do we apply the chain rule for composite functions?
2. What is the geometric meaning of a tangent line?
3. Why does the derivative of $\tan^{-1}(x)$ have the form $\frac{1}{1 + x^2}$?
4. How do we evaluate the inverse trigonometric function $\tan^{-1}(x)$?
5. What are other ways to find the tangent line if the function is not differentiable?

Tip: Always remember to evaluate the slope at the given point when using the point-slope formula for tangent lines.