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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![### Calculus Problem – Logarithmic Differentiation
**Problem:**
12. If \( y = x^{\ln x} \), then \( y' \) is...
(Hint: Use logarithmic differentiation.)
**Solution:**
To solve this problem, we'll use the technique of logarithmic differentiation. Logarithmic differentiation is especially useful when you have a function in the form of one variable raised to another variable. Here, we proceed with the following steps:
1. **Take the Natural Logarithm of Both Sides:**
Given:
\[
y = x^{\ln x}
\]
Taking the natural logarithm (ln) of both sides:
\[
\ln y = \ln (x^{\ln x})
\]
2. **Simplify Using Logarithm Properties:**
Use the property of logarithms that \(\ln(a^b) = b \ln a\):
\[
\ln y = \ln x \cdot \ln x
\]
\[
\ln y = (\ln x)^2
\]
3. **Differentiate Both Sides with Respect to \(x\):**
Recall that \(\frac{d}{dx} (\ln y) = \frac{1}{y} \cdot y'\):
\[
\frac{1}{y} \cdot y' = \frac{d}{dx}((\ln x)^2)
\]
Using the chain rule on the right-hand side:
\[
\frac{1}{y} \cdot y' = 2 \ln x \cdot \frac{1}{x}
\]
\[
\frac{1}{y} \cdot y' = \frac{2 \ln x}{x}
\]
4. **Solve for \( y' \):**
Multiply both sides by \( y \):
\[
y' = y \cdot \frac{2 \ln x}{x}
\]
5. **Substitute \( y \) Back into the Expression:**
Recall that \( y = x^{\ln x} \):
\[
y' = x^{\ln x} \cdot \frac{2 \ln x}{x}
\]
Simplify further:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa451162e-14ac-4cfd-a2c6-1eae7f4a5f1a%2F399505d5-d2d9-48b1-a13e-a20790f892e1%2Fv6wq0q_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus Problem – Logarithmic Differentiation
**Problem:**
12. If \( y = x^{\ln x} \), then \( y' \) is...
(Hint: Use logarithmic differentiation.)
**Solution:**
To solve this problem, we'll use the technique of logarithmic differentiation. Logarithmic differentiation is especially useful when you have a function in the form of one variable raised to another variable. Here, we proceed with the following steps:
1. **Take the Natural Logarithm of Both Sides:**
Given:
\[
y = x^{\ln x}
\]
Taking the natural logarithm (ln) of both sides:
\[
\ln y = \ln (x^{\ln x})
\]
2. **Simplify Using Logarithm Properties:**
Use the property of logarithms that \(\ln(a^b) = b \ln a\):
\[
\ln y = \ln x \cdot \ln x
\]
\[
\ln y = (\ln x)^2
\]
3. **Differentiate Both Sides with Respect to \(x\):**
Recall that \(\frac{d}{dx} (\ln y) = \frac{1}{y} \cdot y'\):
\[
\frac{1}{y} \cdot y' = \frac{d}{dx}((\ln x)^2)
\]
Using the chain rule on the right-hand side:
\[
\frac{1}{y} \cdot y' = 2 \ln x \cdot \frac{1}{x}
\]
\[
\frac{1}{y} \cdot y' = \frac{2 \ln x}{x}
\]
4. **Solve for \( y' \):**
Multiply both sides by \( y \):
\[
y' = y \cdot \frac{2 \ln x}{x}
\]
5. **Substitute \( y \) Back into the Expression:**
Recall that \( y = x^{\ln x} \):
\[
y' = x^{\ln x} \cdot \frac{2 \ln x}{x}
\]
Simplify further:
\[
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