Find an equation for the line tangent to the graph of y equals tangent Superscript negative 1 Baseline left parenthesis one fourth x right parenthesis
 at the point left parenthesis 4 comma StartFraction pi Over 4 EndFraction right parenthesis
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Part 1
What is the tangent line to y equals tangent Superscript negative 1 Baseline left parenthesis one fourth x right parenthesis
 at the point left parenthesis 4 comma StartFraction pi Over 4 EndFraction right parenthesis
​?

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.

Find an equation for the line tangent to the graph of y equals tangent Superscript negative 1 Baseline left parenthesis one fourth x right parenthesis at the point left parenthesis 4 comma StartFraction pi Over 4 EndFraction right parenthesis.

Learn how to find the equation of the tangent line to the function y = arctan(1/4x) at the point (4, π/4). This detailed solution involves applying calculus concepts such as differentiation, the chain rule, and the point-slope formula, suitable for advanced high school and college-level students.

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