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8. Spom INSTRUCTIONS: Find the line tangent to f(x) = sin(ln(x)) at point (1,0)

8. Spom INSTRUCTIONS: Find the line tangent to f(x) = sin(ln(x)) at point (1,0)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Instructions for Problem #8:** Find the line tangent to \( f(x) = \sin(\ln(x)) \) at point \( (1, 0) \). **Guidance for Solving the Problem:** 1. **Understand the Function:** You are working with a composite function where you take the sine of the natural logarithm of \( x \). 2. **Find the Derivative:** Use the chain rule to differentiate \( f(x) \). - Let \( u = \ln(x) \), so \( f(x) = \sin(u) \). - \( f'(x) = \cos(u) \cdot u'(x) \). - Since \( u = \ln(x) \), \( u'(x) = \frac{1}{x} \). - Therefore, \( f'(x) = \cos(\ln(x)) \cdot \frac{1}{x} \). 3. **Evaluate the Derivative at \( x = 1 \):** - Substitute \( x = 1 \) into the derivative: \( f'(1) = \cos(\ln(1)) \cdot \frac{1}{1} \). - Since \( \ln(1) = 0 \), \( \cos(0) = 1 \), so \( f'(1) = 1 \). 4. **Equation of the Tangent Line:** - Use the point-slope form of the line: \( y - y_1 = m(x - x_1) \). - Here, \( m = 1 \), \( (x_1, y_1) = (1, 0) \). - So, the equation is \( y - 0 = 1(x - 1) \). - Simplifies to \( y = x - 1 \). **Conclusion:** The equation of the tangent line to \( f(x) = \sin(\ln(x)) \) at the point \( (1, 0) \) is \( y = x - 1 \).
Transcribed Image Text:**Instructions for Problem #8:** Find the line tangent to \( f(x) = \sin(\ln(x)) \) at point \( (1, 0) \). **Guidance for Solving the Problem:** 1. **Understand the Function:** You are working with a composite function where you take the sine of the natural logarithm of \( x \). 2. **Find the Derivative:** Use the chain rule to differentiate \( f(x) \). - Let \( u = \ln(x) \), so \( f(x) = \sin(u) \). - \( f'(x) = \cos(u) \cdot u'(x) \). - Since \( u = \ln(x) \), \( u'(x) = \frac{1}{x} \). - Therefore, \( f'(x) = \cos(\ln(x)) \cdot \frac{1}{x} \). 3. **Evaluate the Derivative at \( x = 1 \):** - Substitute \( x = 1 \) into the derivative: \( f'(1) = \cos(\ln(1)) \cdot \frac{1}{1} \). - Since \( \ln(1) = 0 \), \( \cos(0) = 1 \), so \( f'(1) = 1 \). 4. **Equation of the Tangent Line:** - Use the point-slope form of the line: \( y - y_1 = m(x - x_1) \). - Here, \( m = 1 \), \( (x_1, y_1) = (1, 0) \). - So, the equation is \( y - 0 = 1(x - 1) \). - Simplifies to \( y = x - 1 \). **Conclusion:** The equation of the tangent line to \( f(x) = \sin(\ln(x)) \) at the point \( (1, 0) \) is \( y = x - 1 \).
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