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
![**Compute the derivative.**
\[ y = \cos \left( te^{-6t} \right) \]
(Express numbers in exact form. Use symbolic notation and fractions where needed.)
\[ y' = \underline{\hspace{6cm}} \]
**Explanation (For Educational Purposes):**
In this problem, we are asked to compute the derivative of the function \( y = \cos \left( te^{-6t} \right) \).
To solve this, we need to apply the chain rule and the product rule of differentiation, as the argument of the cosine function is itself a product of two functions of \( t \).
1. **Identify the outer function and its derivative:**
- Outer function: \( u = \cos(u) \)
- Derivative of the outer function: \( \frac{d}{du} \cos(u) = -\sin(u) \)
2. **Identify the inner function and its derivative:**
- Inner function: \( v = te^{-6t} \)
- To differentiate this product, we'll use the product rule: \( v = t \cdot e^{-6t} \)
3. **Apply the product rule to the inner function:**
\[
\frac{d}{dt} \left( te^{-6t} \right) = e^{-6t} + t \cdot \frac{d}{dt}(e^{-6t})
\]
- Using the chain rule, the derivative of \( e^{-6t} \) is \( e^{-6t} \cdot (-6) = -6e^{-6t} \)
- Therefore,
\[
\frac{d}{dt} \left( te^{-6t} \right) = e^{-6t} + t(-6e^{-6t}) = e^{-6t} - 6te^{-6t}
\]
4. **Combine these results using the chain rule:**
\[
y' = -\sin(te^{-6t}) \cdot \left( e^{-6t} - 6te^{-6t} \right)
\]
- Simplifying, we get:
\[
y' = -\sin(te^{-6t}) \cdot e^{-6t} \left( 1 - 6t \right)
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F575d721b-26eb-4db6-af93-50c207cd3fab%2F92563587-85c7-4708-ae12-4fe88f10dc71%2Fn1ydd1o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Compute the derivative.**
\[ y = \cos \left( te^{-6t} \right) \]
(Express numbers in exact form. Use symbolic notation and fractions where needed.)
\[ y' = \underline{\hspace{6cm}} \]
**Explanation (For Educational Purposes):**
In this problem, we are asked to compute the derivative of the function \( y = \cos \left( te^{-6t} \right) \).
To solve this, we need to apply the chain rule and the product rule of differentiation, as the argument of the cosine function is itself a product of two functions of \( t \).
1. **Identify the outer function and its derivative:**
- Outer function: \( u = \cos(u) \)
- Derivative of the outer function: \( \frac{d}{du} \cos(u) = -\sin(u) \)
2. **Identify the inner function and its derivative:**
- Inner function: \( v = te^{-6t} \)
- To differentiate this product, we'll use the product rule: \( v = t \cdot e^{-6t} \)
3. **Apply the product rule to the inner function:**
\[
\frac{d}{dt} \left( te^{-6t} \right) = e^{-6t} + t \cdot \frac{d}{dt}(e^{-6t})
\]
- Using the chain rule, the derivative of \( e^{-6t} \) is \( e^{-6t} \cdot (-6) = -6e^{-6t} \)
- Therefore,
\[
\frac{d}{dt} \left( te^{-6t} \right) = e^{-6t} + t(-6e^{-6t}) = e^{-6t} - 6te^{-6t}
\]
4. **Combine these results using the chain rule:**
\[
y' = -\sin(te^{-6t}) \cdot \left( e^{-6t} - 6te^{-6t} \right)
\]
- Simplifying, we get:
\[
y' = -\sin(te^{-6t}) \cdot e^{-6t} \left( 1 - 6t \right)
\]
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