**Topic: Logarithmic Differentiation** To find \(\frac{dy}{dx}\) using logarithmic differentiation, consider the given function: \[ y = \sqrt{\frac{x^2 + 1}{e^{8x} \sec x}} \] ### Steps for Logarithmic Differentiation: 1. **Take Natural Logarithms:** Apply the natural logarithm to both sides of the equation: \[ \ln y = \ln \left( \sqrt{\frac{x^2 + 1}{e^{8x} \sec x}} \right) \] 2. **Simplify Using Logarithm Rules:** Use the property \(\ln \sqrt{a} = \frac{1}{2} \ln a\) and the quotient rule for logarithms: \[ \ln y = \frac{1}{2} \left( \ln(x^2 + 1) - \ln(e^{8x} \sec x) \right) \] \[ = \frac{1}{2} (\ln(x^2 + 1) - 8x - \ln(\sec x)) \] 3. **Differentiate Both Sides with Respect to \(x\):** \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( \frac{2x}{x^2 + 1} - 8 - \tan x \right) \] 4. **Solve for \(\frac{dy}{dx}\):** \[ \frac{dy}{dx} = y \cdot \frac{1}{2} \left( \frac{2x}{x^2 + 1} - 8 - \tan x \right) \] 5. **Substitute Back \(y\):** \[ \frac{dy}{dx} = \sqrt{\frac{x^2 + 1}{e^{8x} \sec x}} \cdot \frac{1}{2} \left( \frac{2x}{x^2 + 1} - 8 - \tan x \right) \] This completes the steps for finding the derivative of the given function using logarithmic differentiation.
**Topic: Logarithmic Differentiation** To find \(\frac{dy}{dx}\) using logarithmic differentiation, consider the given function: \[ y = \sqrt{\frac{x^2 + 1}{e^{8x} \sec x}} \] ### Steps for Logarithmic Differentiation: 1. **Take Natural Logarithms:** Apply the natural logarithm to both sides of the equation: \[ \ln y = \ln \left( \sqrt{\frac{x^2 + 1}{e^{8x} \sec x}} \right) \] 2. **Simplify Using Logarithm Rules:** Use the property \(\ln \sqrt{a} = \frac{1}{2} \ln a\) and the quotient rule for logarithms: \[ \ln y = \frac{1}{2} \left( \ln(x^2 + 1) - \ln(e^{8x} \sec x) \right) \] \[ = \frac{1}{2} (\ln(x^2 + 1) - 8x - \ln(\sec x)) \] 3. **Differentiate Both Sides with Respect to \(x\):** \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( \frac{2x}{x^2 + 1} - 8 - \tan x \right) \] 4. **Solve for \(\frac{dy}{dx}\):** \[ \frac{dy}{dx} = y \cdot \frac{1}{2} \left( \frac{2x}{x^2 + 1} - 8 - \tan x \right) \] 5. **Substitute Back \(y\):** \[ \frac{dy}{dx} = \sqrt{\frac{x^2 + 1}{e^{8x} \sec x}} \cdot \frac{1}{2} \left( \frac{2x}{x^2 + 1} - 8 - \tan x \right) \] This completes the steps for finding the derivative of the given function using logarithmic differentiation.
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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