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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![**Problem Statement:**
Use logarithmic differentiation to find \(\frac{dy}{dx}\) if \(y = x^{3x}\).
**Method:**
Logarithmic differentiation is a technique that simplifies the process of differentiating functions by applying the logarithm first. This method is especially useful when dealing with powers where both the base and the exponent are variables, as in this case with \(y = x^{3x}\).
**Steps to Solve:**
1. **Take the Natural Logarithm of Both Sides:**
Start by taking the natural logarithm of both sides of the equation:
\[
\ln y = \ln(x^{3x})
\]
2. **Use Logarithmic Properties:**
Use the property of logarithms that states \(\ln(a^b) = b \ln a\):
\[
\ln y = 3x \ln x
\]
3. **Differentiate Both Sides:**
Differentiate implicitly with respect to \(x\):
\[
\frac{d}{dx}(\ln y) = \frac{d}{dx}(3x \ln x)
\]
Using the chain rule on the left side \(\frac{1}{y} \frac{dy}{dx}\) and applying the product and chain rules on the right side:
\[
\frac{1}{y} \frac{dy}{dx} = \left(3 \ln x + 3x \cdot \frac{1}{x}\right)
\]
Simplify the right side:
\[
\frac{1}{y} \frac{dy}{dx} = 3 \ln x + 3
\]
4. **Solve for \(\frac{dy}{dx}\):**
Multiply through by \(y\) to solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = y(3 \ln x + 3)
\]
5. **Substitute Back for \(y\):**
Since \(y = x^{3x}\), substitute back:
\[
\frac{dy}{dx} = x^{3x}(3 \ln x + 3)
\]
**Conclusion:**
Thus, by using logarithmic differentiation, the derivative of \(y = x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdeff5ce5-c161-409f-a0ee-bdeaef8b0965%2F1fd1aa4a-dea0-438f-8efd-e63900309455%2Fmjqd9no_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Use logarithmic differentiation to find \(\frac{dy}{dx}\) if \(y = x^{3x}\).
**Method:**
Logarithmic differentiation is a technique that simplifies the process of differentiating functions by applying the logarithm first. This method is especially useful when dealing with powers where both the base and the exponent are variables, as in this case with \(y = x^{3x}\).
**Steps to Solve:**
1. **Take the Natural Logarithm of Both Sides:**
Start by taking the natural logarithm of both sides of the equation:
\[
\ln y = \ln(x^{3x})
\]
2. **Use Logarithmic Properties:**
Use the property of logarithms that states \(\ln(a^b) = b \ln a\):
\[
\ln y = 3x \ln x
\]
3. **Differentiate Both Sides:**
Differentiate implicitly with respect to \(x\):
\[
\frac{d}{dx}(\ln y) = \frac{d}{dx}(3x \ln x)
\]
Using the chain rule on the left side \(\frac{1}{y} \frac{dy}{dx}\) and applying the product and chain rules on the right side:
\[
\frac{1}{y} \frac{dy}{dx} = \left(3 \ln x + 3x \cdot \frac{1}{x}\right)
\]
Simplify the right side:
\[
\frac{1}{y} \frac{dy}{dx} = 3 \ln x + 3
\]
4. **Solve for \(\frac{dy}{dx}\):**
Multiply through by \(y\) to solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = y(3 \ln x + 3)
\]
5. **Substitute Back for \(y\):**
Since \(y = x^{3x}\), substitute back:
\[
\frac{dy}{dx} = x^{3x}(3 \ln x + 3)
\]
**Conclusion:**
Thus, by using logarithmic differentiation, the derivative of \(y = x
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