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 - Finding Derivatives of Hyperbolic Functions**
In this exercise, we aim to find the derivatives of the given hyperbolic functions.
1. Find each of the following derivatives.
a. \(\frac{d}{dx} \left[ \cosh(2x^2 - 4) \right]\)
b. \(\frac{d}{dx} \left[ \tanh(4e^{3x+1}) \right]\)
### Steps to Solve:
#### Part (a):
To find \(\frac{d}{dx} \left[ \cosh(2x^2 - 4) \right]\), we will use the chain rule.
1. **Derivative of the Outer Function**: The derivative of \(\cosh(u)\) is \(\sinh(u)\).
2. **Derivative of the Inner Function**: Let \(u = 2x^2 - 4\). The derivative of \(u\) with respect to \(x\) is \(4x\).
Putting it together:
\[
\frac{d}{dx} \left[ \cosh(2x^2 - 4) \right] = \sinh(2x^2 - 4) \cdot 4x
\]
#### Part (b):
To find \(\frac{d}{dx} \left[ \tanh(4e^{3x+1}) \right]\), we will use the chain rule again.
1. **Derivative of the Outer Function**: The derivative of \(\tanh(u)\) is \(\text{sech}^2(u)\).
2. **Derivative of the Inner Function**: Let \(u = 4e^{3x+1}\). The derivative of \(u\) with respect to \(x\) is:
\[
\frac{d}{dx} [4e^{3x+1}] = 4 \cdot e^{3x+1} \cdot 3 = 12e^{3x+1}
\]
Putting it together:
\[
\frac{d}{dx} \left[ \tanh(4e^{3x+1}) \right] = \text{sech}^2(4e^{3x+1}) \cdot 12e^{3x+1}
\]
By following these](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a885dd9-3ed1-489d-bae2-5a33154ef1ff%2Fc305cccd-6d67-4f9c-a297-d3170cfdff5e%2F5rknpyw_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculus - Finding Derivatives of Hyperbolic Functions**
In this exercise, we aim to find the derivatives of the given hyperbolic functions.
1. Find each of the following derivatives.
a. \(\frac{d}{dx} \left[ \cosh(2x^2 - 4) \right]\)
b. \(\frac{d}{dx} \left[ \tanh(4e^{3x+1}) \right]\)
### Steps to Solve:
#### Part (a):
To find \(\frac{d}{dx} \left[ \cosh(2x^2 - 4) \right]\), we will use the chain rule.
1. **Derivative of the Outer Function**: The derivative of \(\cosh(u)\) is \(\sinh(u)\).
2. **Derivative of the Inner Function**: Let \(u = 2x^2 - 4\). The derivative of \(u\) with respect to \(x\) is \(4x\).
Putting it together:
\[
\frac{d}{dx} \left[ \cosh(2x^2 - 4) \right] = \sinh(2x^2 - 4) \cdot 4x
\]
#### Part (b):
To find \(\frac{d}{dx} \left[ \tanh(4e^{3x+1}) \right]\), we will use the chain rule again.
1. **Derivative of the Outer Function**: The derivative of \(\tanh(u)\) is \(\text{sech}^2(u)\).
2. **Derivative of the Inner Function**: Let \(u = 4e^{3x+1}\). The derivative of \(u\) with respect to \(x\) is:
\[
\frac{d}{dx} [4e^{3x+1}] = 4 \cdot e^{3x+1} \cdot 3 = 12e^{3x+1}
\]
Putting it together:
\[
\frac{d}{dx} \left[ \tanh(4e^{3x+1}) \right] = \text{sech}^2(4e^{3x+1}) \cdot 12e^{3x+1}
\]
By following these
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