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
Problem 1RQ
Related questions
Question
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 \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2a8966ca-731d-4724-8043-4e7365cac85e%2F58e7754c-118d-4eef-99ad-400bab55e2e5%2Firua5rr_processed.png&w=3840&q=75)
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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