Find the slope of the tangent line to the graph of y = tan¬1 () at x = 4

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Find the slope of the tangent to the graph at a given point?

**Problem Statement:**

Find the slope of the tangent line to the graph of \( y = \tan^{-1} \left( \frac{x}{4} \right) \) at \( x = 4 \).

**Solution:**

To find the slope of the tangent line, we need to compute the derivative of the function with respect to \( x \), and then evaluate it at \( x = 4 \).

1. **Differentiate the Function:**

   The function is \( y = \tan^{-1} \left( \frac{x}{4} \right) \).

   Use the chain rule to differentiate:
   \[
   \frac{dy}{dx} = \frac{1}{1 + \left( \frac{x}{4} \right)^2} \cdot \frac{d}{dx} \left(\frac{x}{4}\right)
   \]

   The derivative of \( \frac{x}{4} \) with respect to \( x \) is \( \frac{1}{4} \).

   Therefore,
   \[
   \frac{dy}{dx} = \frac{1}{1 + \left( \frac{x}{4} \right)^2} \cdot \frac{1}{4}
   \]

   Simplify the expression:
   \[
   \frac{dy}{dx} = \frac{1}{4} \cdot \frac{1}{1 + \frac{x^2}{16}}
   \]

   Simplify further:
   \[
   \frac{dy}{dx} = \frac{1}{4 + \frac{x^2}{4}}
   \]

   Combine the denominators:
   \[
   \frac{dy}{dx} = \frac{4}{16 + x^2}
   \]

2. **Evaluate the Derivative at \( x = 4 \):**

   Substitute \( x = 4 \) into the derivative:
   \[
   \frac{dy}{dx} = \frac{4}{16 + 4^2} = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8}
   \]

**Conclusion:**

The slope of the tangent line to the graph of \( y = \tan^{-1} \left( \frac{x}{4} \right) \) at \(
Transcribed Image Text:**Problem Statement:** Find the slope of the tangent line to the graph of \( y = \tan^{-1} \left( \frac{x}{4} \right) \) at \( x = 4 \). **Solution:** To find the slope of the tangent line, we need to compute the derivative of the function with respect to \( x \), and then evaluate it at \( x = 4 \). 1. **Differentiate the Function:** The function is \( y = \tan^{-1} \left( \frac{x}{4} \right) \). Use the chain rule to differentiate: \[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{x}{4} \right)^2} \cdot \frac{d}{dx} \left(\frac{x}{4}\right) \] The derivative of \( \frac{x}{4} \) with respect to \( x \) is \( \frac{1}{4} \). Therefore, \[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{x}{4} \right)^2} \cdot \frac{1}{4} \] Simplify the expression: \[ \frac{dy}{dx} = \frac{1}{4} \cdot \frac{1}{1 + \frac{x^2}{16}} \] Simplify further: \[ \frac{dy}{dx} = \frac{1}{4 + \frac{x^2}{4}} \] Combine the denominators: \[ \frac{dy}{dx} = \frac{4}{16 + x^2} \] 2. **Evaluate the Derivative at \( x = 4 \):** Substitute \( x = 4 \) into the derivative: \[ \frac{dy}{dx} = \frac{4}{16 + 4^2} = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8} \] **Conclusion:** The slope of the tangent line to the graph of \( y = \tan^{-1} \left( \frac{x}{4} \right) \) at \(
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