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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![**Differentiate the function.**
Given the function:
\[ g(t) = \ln\left(\frac{t(t^2 + 1)^8}{\sqrt[3]{4t - 1}}\right) \]
Find the derivative \( g'(t) \):
\[ g'(t) = \boxed{\phantom{g'(t) =}} \]
**Explanation of the Function and Derivative Process:**
The function to differentiate involves a natural logarithm \(\ln\) of a quotient. The expression inside the logarithm includes both an exponential term \((t^2 + 1)^8\) and a cube root \(\sqrt[3]{4t - 1}\). Differentiation of logarithmic functions typically requires using the chain rule and the properties of logarithms:
1. **Rewrite the expression using properties of logarithms:**
- \(\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)\)
- For \(A = t(t^2 + 1)^8\) and \(B = \sqrt[3]{4t - 1}\), this becomes:
\[ \ln\left(t(t^2 + 1)^8\right) - \ln\left(\sqrt[3]{4t - 1}\right) \]
2. **Break down further using logarithm properties:**
- \(\ln(t) + 8\ln(t^2 + 1)\)
- \(\frac{1}{3}\ln(4t - 1)\)
3. **Differentiate each term separately using basic differentiation techniques:**
- The derivative of \(\ln(t)\) is \(\frac{1}{t}\).
- The derivative of \(8\ln(t^2 + 1)\) involves the chain rule: \(8 \cdot \frac{1}{t^2 + 1} \cdot 2t = \frac{16t}{t^2 + 1}\).
- The derivative of \(\frac{1}{3}\ln(4t - 1)\) involves the chain rule: \(\frac{1}{3}\cdot \frac{1}{4t - 1} \cdot 4 = \frac{4}{3(4t - 1)}\).
The](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2F058c662f-6d66-4521-b850-b3e14f6b8fa2%2Fl5494s7_processed.png&w=3840&q=75)
Transcribed Image Text:**Differentiate the function.**
Given the function:
\[ g(t) = \ln\left(\frac{t(t^2 + 1)^8}{\sqrt[3]{4t - 1}}\right) \]
Find the derivative \( g'(t) \):
\[ g'(t) = \boxed{\phantom{g'(t) =}} \]
**Explanation of the Function and Derivative Process:**
The function to differentiate involves a natural logarithm \(\ln\) of a quotient. The expression inside the logarithm includes both an exponential term \((t^2 + 1)^8\) and a cube root \(\sqrt[3]{4t - 1}\). Differentiation of logarithmic functions typically requires using the chain rule and the properties of logarithms:
1. **Rewrite the expression using properties of logarithms:**
- \(\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)\)
- For \(A = t(t^2 + 1)^8\) and \(B = \sqrt[3]{4t - 1}\), this becomes:
\[ \ln\left(t(t^2 + 1)^8\right) - \ln\left(\sqrt[3]{4t - 1}\right) \]
2. **Break down further using logarithm properties:**
- \(\ln(t) + 8\ln(t^2 + 1)\)
- \(\frac{1}{3}\ln(4t - 1)\)
3. **Differentiate each term separately using basic differentiation techniques:**
- The derivative of \(\ln(t)\) is \(\frac{1}{t}\).
- The derivative of \(8\ln(t^2 + 1)\) involves the chain rule: \(8 \cdot \frac{1}{t^2 + 1} \cdot 2t = \frac{16t}{t^2 + 1}\).
- The derivative of \(\frac{1}{3}\ln(4t - 1)\) involves the chain rule: \(\frac{1}{3}\cdot \frac{1}{4t - 1} \cdot 4 = \frac{4}{3(4t - 1)}\).
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