Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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I need help finding the 2nd derivative of y. I believe I'm on the right track but I'm not sure how to continue.
![**Objective:** Find the first and second derivatives, \( y' \) and \( y'' \).
**Given Function:**
\[ y = \sqrt{\sin(x)} \]
**First Derivative (\( y' \)):**
The derivative is calculated as:
\[ y' = \frac{\cos(x)}{2\sqrt{\sin(x)}} \]
**Second Derivative (\( y'' \)):**
The second derivative is not provided and is expected to be calculated based on \( y' \).
**Explanation:**
To find the first derivative of \( y = \sqrt{\sin(x)} \), you can use the chain rule. The process involves differentiating the outer and inner functions.
1. The outer function is \( u^{1/2} \), where \( u = \sin(x) \), giving the derivative as \( \frac{1}{2}u^{-1/2} \).
2. The inner function is \( u = \sin(x) \), whose derivative is \( \cos(x) \).
Thus, combining these using the chain rule results in:
\[ y' = \frac{1}{2} (\sin(x))^{-1/2} \cdot \cos(x) = \frac{\cos(x)}{2\sqrt{\sin(x)}} \]
The second derivative \( y'' \) requires further differentiation of \( y' \).](https://content.bartleby.com/qna-images/question/3eb03a80-7cae-4d33-b228-e2e0361c370e/ffae9de3-591e-4f79-9f70-87014a586ceb/hpft2zq_processed.png)
Transcribed Image Text:**Objective:** Find the first and second derivatives, \( y' \) and \( y'' \).
**Given Function:**
\[ y = \sqrt{\sin(x)} \]
**First Derivative (\( y' \)):**
The derivative is calculated as:
\[ y' = \frac{\cos(x)}{2\sqrt{\sin(x)}} \]
**Second Derivative (\( y'' \)):**
The second derivative is not provided and is expected to be calculated based on \( y' \).
**Explanation:**
To find the first derivative of \( y = \sqrt{\sin(x)} \), you can use the chain rule. The process involves differentiating the outer and inner functions.
1. The outer function is \( u^{1/2} \), where \( u = \sin(x) \), giving the derivative as \( \frac{1}{2}u^{-1/2} \).
2. The inner function is \( u = \sin(x) \), whose derivative is \( \cos(x) \).
Thus, combining these using the chain rule results in:
\[ y' = \frac{1}{2} (\sin(x))^{-1/2} \cdot \cos(x) = \frac{\cos(x)}{2\sqrt{\sin(x)}} \]
The second derivative \( y'' \) requires further differentiation of \( y' \).
![This image contains the step-by-step differentiation of a mathematical function using the quotient rule. Below is a detailed transcription suitable for an educational website:
---
**Quotient Rule Differentiation**
Given:
\[ y' = \frac{\cos(x)}{2\sqrt{\sin(x)}} \]
To differentiate using the quotient rule, we follow these steps:
1. **Identify the Functions:**
- Let \( f(x) = \cos(x) \) and \( g(x) = 2\sqrt{\sin(x)} \).
- The derivative using the quotient rule is \(\frac{gf' - fg'}{g^2}\).
2. **Apply the Quotient Rule:**
\[
y'' = \frac{2\sqrt{\sin(x)} \cdot \frac{d}{dx}[\cos(x)] - \cos(x) \cdot \frac{d}{dx}[2\sqrt{\sin(x)}]}{(2\sqrt{\sin(x)})^2}
\]
3. **Calculate the Derivatives:**
- Derivative of \(\cos(x)\) is \( -\sin(x) \).
- Differentiate \(2\sqrt{\sin(x)}\) using the chain rule:
\[
2 \cdot \frac{d}{dx}\left[ \sin(x)^{1/2} \right] = 2 \cdot \frac{1}{2} \cdot \sin(x)^{-1/2} \cdot \frac{d}{dx}[\sin(x)]
\]
4. **Substitute and Simplify:**
\[
y'' = \frac{2\sqrt{\sin(x)} \cdot (-\sin(x)) - \cos(x) \cdot 2 \cdot \left(\frac{1}{2} \cdot \sin(x)^{-1/2} \cdot \cos(x)\right)}{(2\sqrt{\sin(x)})^2}
\]
5. **Final Simplified Expression:**
\[
y'' = \frac{2\sqrt{\sin(x)} \cdot (-\sin(x)) - \cos(x) \cdot \sin(x)^{-1/2} \cdot \cos(x)}{(2\sqrt{\sin(x)})^2}](https://content.bartleby.com/qna-images/question/3eb03a80-7cae-4d33-b228-e2e0361c370e/ffae9de3-591e-4f79-9f70-87014a586ceb/w0hxfs_processed.png)
Transcribed Image Text:This image contains the step-by-step differentiation of a mathematical function using the quotient rule. Below is a detailed transcription suitable for an educational website:
---
**Quotient Rule Differentiation**
Given:
\[ y' = \frac{\cos(x)}{2\sqrt{\sin(x)}} \]
To differentiate using the quotient rule, we follow these steps:
1. **Identify the Functions:**
- Let \( f(x) = \cos(x) \) and \( g(x) = 2\sqrt{\sin(x)} \).
- The derivative using the quotient rule is \(\frac{gf' - fg'}{g^2}\).
2. **Apply the Quotient Rule:**
\[
y'' = \frac{2\sqrt{\sin(x)} \cdot \frac{d}{dx}[\cos(x)] - \cos(x) \cdot \frac{d}{dx}[2\sqrt{\sin(x)}]}{(2\sqrt{\sin(x)})^2}
\]
3. **Calculate the Derivatives:**
- Derivative of \(\cos(x)\) is \( -\sin(x) \).
- Differentiate \(2\sqrt{\sin(x)}\) using the chain rule:
\[
2 \cdot \frac{d}{dx}\left[ \sin(x)^{1/2} \right] = 2 \cdot \frac{1}{2} \cdot \sin(x)^{-1/2} \cdot \frac{d}{dx}[\sin(x)]
\]
4. **Substitute and Simplify:**
\[
y'' = \frac{2\sqrt{\sin(x)} \cdot (-\sin(x)) - \cos(x) \cdot 2 \cdot \left(\frac{1}{2} \cdot \sin(x)^{-1/2} \cdot \cos(x)\right)}{(2\sqrt{\sin(x)})^2}
\]
5. **Final Simplified Expression:**
\[
y'' = \frac{2\sqrt{\sin(x)} \cdot (-\sin(x)) - \cos(x) \cdot \sin(x)^{-1/2} \cdot \cos(x)}{(2\sqrt{\sin(x)})^2}
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Could you explain the last two parts to me? How do you go from sin(x)3/2 to sin2(x)?
And where did square root of sin(x) go? Did you somehow move it to the denominator?
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Follow-up Question
Could you explain the last two parts to me? How do you go from sin(x)3/2 to sin2(x)?
And where did square root of sin(x) go? Did you somehow move it to the denominator?
Solution
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