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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![**Problem Statement:**
Find \( dy \) for \( y = -5 \sec \left(\frac{5x^8}{8}\right) \).
**Solution:**
To find the derivative of the function \( y \) with respect to \( x \), we'll need to apply the chain rule. The function is of the form \( y = -5 \sec(u) \) where \( u = \frac{5x^8}{8} \).
**Steps:**
1. **Differentiate the outer function**: The derivative of \( -5 \sec(u) \) with respect to \( u \) is \( -5 \sec(u) \tan(u) \).
2. **Differentiate the inner function**: The derivative of \( u = \frac{5x^8}{8} \) with respect to \( x \) is \( \frac{40x^7}{8} \) or \( 5x^7 \).
3. **Apply the chain rule**: Combine these using the chain rule: \( \frac{dy}{dx} = -5 \sec(u) \tan(u) \cdot 5x^7 \).
Therefore, the expression for \( dy \) is:
\[ dy = \left(-25 \sec\left(\frac{5x^8}{8}\right) \tan\left(\frac{5x^8}{8}\right) \cdot x^7\right) dx \]
This gives the final expression for the differential \( dy \) based on the given function \( y \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeadb183-1175-48ee-be1d-be4cbd36e1e9%2Ffc3515b9-9b3b-46b4-885a-ffacf90017f5%2Fhd4q9x_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \( dy \) for \( y = -5 \sec \left(\frac{5x^8}{8}\right) \).
**Solution:**
To find the derivative of the function \( y \) with respect to \( x \), we'll need to apply the chain rule. The function is of the form \( y = -5 \sec(u) \) where \( u = \frac{5x^8}{8} \).
**Steps:**
1. **Differentiate the outer function**: The derivative of \( -5 \sec(u) \) with respect to \( u \) is \( -5 \sec(u) \tan(u) \).
2. **Differentiate the inner function**: The derivative of \( u = \frac{5x^8}{8} \) with respect to \( x \) is \( \frac{40x^7}{8} \) or \( 5x^7 \).
3. **Apply the chain rule**: Combine these using the chain rule: \( \frac{dy}{dx} = -5 \sec(u) \tan(u) \cdot 5x^7 \).
Therefore, the expression for \( dy \) is:
\[ dy = \left(-25 \sec\left(\frac{5x^8}{8}\right) \tan\left(\frac{5x^8}{8}\right) \cdot x^7\right) dx \]
This gives the final expression for the differential \( dy \) based on the given function \( y \).
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