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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Question
![To solve the problem of finding \(\frac{dy}{dt}\) given the function \(y = \ln \left( \frac{t^2 + 9}{\sqrt{8 - t}} \right)\), you will need to apply the chain rule and quotient rule from calculus.
### Solution Steps:
1. **Identify Parts of the Function:**
- Let \(u = t^2 + 9\) and \(v = \sqrt{8 - t}\).
- The function becomes \(y = \ln \left( \frac{u}{v} \right)\).
2. **Apply Logarithmic Differentiation:**
- The derivative of \(\ln \left( \frac{u}{v} \right)\) is \(\frac{1}{\left( \frac{u}{v} \right)} \cdot \left( \frac{du/dt \cdot v - u \cdot dv/dt}{v^2} \right)\).
3. **Find Derivatives of \(u\) and \(v\):**
- \(du/dt = 2t\).
- \(dv/dt\) requires implicit differentiation:
- \(v = (8-t)^{1/2}\), so \(dv/dt = -\frac{1}{2}(8-t)^{-1/2}\).
4. **Substitute into the Derivatives:**
- Compute the derivative \(\frac{du}{v^2} = \frac{2t}{(8-t)}\).
- Compute the second part: \(- \frac{u \cdot \frac{1}{2}(8-t)^{-3/2}}{v^2} = - \frac{(t^2 + 9)}{2(8-t)^{-3/2}}\).
By solving these steps, you'll find the solution for \(\frac{dy}{dt}\). The crucial aspect is correctly applying derivative rules and simplifying the expressions. Once calculated, substitute the values back to get the explicit derivative \(\frac{dy}{dt} = \text{[your solution]}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F246f6db2-1dc3-454a-8242-1f5f62a81278%2Fbb8bf1cd-9dcd-4596-a588-84cdcd2d2fe9%2Fa8nteiq_processed.png&w=3840&q=75)
Transcribed Image Text:To solve the problem of finding \(\frac{dy}{dt}\) given the function \(y = \ln \left( \frac{t^2 + 9}{\sqrt{8 - t}} \right)\), you will need to apply the chain rule and quotient rule from calculus.
### Solution Steps:
1. **Identify Parts of the Function:**
- Let \(u = t^2 + 9\) and \(v = \sqrt{8 - t}\).
- The function becomes \(y = \ln \left( \frac{u}{v} \right)\).
2. **Apply Logarithmic Differentiation:**
- The derivative of \(\ln \left( \frac{u}{v} \right)\) is \(\frac{1}{\left( \frac{u}{v} \right)} \cdot \left( \frac{du/dt \cdot v - u \cdot dv/dt}{v^2} \right)\).
3. **Find Derivatives of \(u\) and \(v\):**
- \(du/dt = 2t\).
- \(dv/dt\) requires implicit differentiation:
- \(v = (8-t)^{1/2}\), so \(dv/dt = -\frac{1}{2}(8-t)^{-1/2}\).
4. **Substitute into the Derivatives:**
- Compute the derivative \(\frac{du}{v^2} = \frac{2t}{(8-t)}\).
- Compute the second part: \(- \frac{u \cdot \frac{1}{2}(8-t)^{-3/2}}{v^2} = - \frac{(t^2 + 9)}{2(8-t)^{-3/2}}\).
By solving these steps, you'll find the solution for \(\frac{dy}{dt}\). The crucial aspect is correctly applying derivative rules and simplifying the expressions. Once calculated, substitute the values back to get the explicit derivative \(\frac{dy}{dt} = \text{[your solution]}\).
![**Problem Statement:**
Find \(\frac{dy}{dx}\) if \( y = \ln(x^3 \sqrt{x+1}) \).
**Solution:**
\[
\frac{dy}{dx} = \boxed{\phantom{x}}
\]
**Explanation:**
To solve this problem, we need to differentiate the given function \( y = \ln(x^3 \sqrt{x+1}) \) with respect to \( x \).
### Steps for Differentiation:
1. **Use Logarithmic Properties:**
Simplify the expression inside the logarithm using properties of logarithms:
\[
y = \ln(x^3 \cdot (x+1)^{1/2}) = \ln(x^3) + \ln((x+1)^{1/2})
\]
This can be further simplified to:
\[
y = 3\ln(x) + \frac{1}{2}\ln(x+1)
\]
2. **Differentiate:**
Apply the derivative:
- The derivative of \( 3\ln(x) \) is \(\frac{3}{x}\).
- The derivative of \( \frac{1}{2}\ln(x+1) \) is \(\frac{1}{2} \cdot \frac{1}{x+1}\).
Combine the derivatives:
\[
\frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)}
\]
This is the derivative of the function. Fill in the box with the expression obtained above.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F246f6db2-1dc3-454a-8242-1f5f62a81278%2Fbb8bf1cd-9dcd-4596-a588-84cdcd2d2fe9%2Fbx5gt5e_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \(\frac{dy}{dx}\) if \( y = \ln(x^3 \sqrt{x+1}) \).
**Solution:**
\[
\frac{dy}{dx} = \boxed{\phantom{x}}
\]
**Explanation:**
To solve this problem, we need to differentiate the given function \( y = \ln(x^3 \sqrt{x+1}) \) with respect to \( x \).
### Steps for Differentiation:
1. **Use Logarithmic Properties:**
Simplify the expression inside the logarithm using properties of logarithms:
\[
y = \ln(x^3 \cdot (x+1)^{1/2}) = \ln(x^3) + \ln((x+1)^{1/2})
\]
This can be further simplified to:
\[
y = 3\ln(x) + \frac{1}{2}\ln(x+1)
\]
2. **Differentiate:**
Apply the derivative:
- The derivative of \( 3\ln(x) \) is \(\frac{3}{x}\).
- The derivative of \( \frac{1}{2}\ln(x+1) \) is \(\frac{1}{2} \cdot \frac{1}{x+1}\).
Combine the derivatives:
\[
\frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)}
\]
This is the derivative of the function. Fill in the box with the expression obtained above.
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