Assignment 1: Derive the Product Rule formula using the definition of a derivative. Your derivation should be neat and clear so that any Calculus student can follow. After EVERY STEP, write one sentence explaining your logic. The following are a few definitions and tips: (1) Definition of the derivative: lim h→0 f (x+h)-f(x) h (2) Use a "magic trick": -f(x + h)g(x) + f(x + h)g(x) (this expression is = 0, so essentially we did not change anything) (3) Factor out f (x + h) and g(x): -f(x + h)g(x) + f(x + h)g(x) To start you off, using the definition of a derivative: f(x + h)g(x + h) lim h→0 < insert trick > -f(x)g(x) h
Assignment 1: Derive the Product Rule formula using the definition of a derivative. Your derivation should be neat and clear so that any Calculus student can follow. After EVERY STEP, write one sentence explaining your logic. The following are a few definitions and tips: (1) Definition of the derivative: lim h→0 f (x+h)-f(x) h (2) Use a "magic trick": -f(x + h)g(x) + f(x + h)g(x) (this expression is = 0, so essentially we did not change anything) (3) Factor out f (x + h) and g(x): -f(x + h)g(x) + f(x + h)g(x) To start you off, using the definition of a derivative: f(x + h)g(x + h) lim h→0 < insert trick > -f(x)g(x) h
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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![## Derivation of the Product and Quotient Rule
### Introduction
In calculus, we have learned that the derivative of the product and quotient of two differentiable functions can be computed using specific rules.
### The Product Rule
If \( f(x) \) and \( g(x) \) are two differentiable functions, the derivative of their product is given by:
\[ \frac{d}{dx} \left[ f(x)g(x) \right] = f(x)g'(x) + g(x)f'(x) \]
### The Quotient Rule
If \( f(x) \) and \( g(x) \) are two differentiable functions where \( g(x) \neq 0 \), the derivative of their quotient is:
\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2} \]
### Assignment 1: Derive the Product Rule
**Objective:** Derive the product rule formula using the definition of a derivative. Ensure every step is explained clearly for any Calculus student to follow.
#### Steps to Follow
1. **Definition of the Derivative:**
\[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
2. **Use a "magic trick":**
\[ -f(x + h)g(x) + f(x + h)g(x) \]
(This expression is 0, so essentially we did not change anything.)
3. **Factor out \( f(x + h) \) and \( g(x) \):**
\[ -f(x + h)g(x) + f(x + h)g(x) \]
To start you off, using the definition of a derivative:
\[ \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} \]
Now insert the "magic trick":
\[ \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h} \]
Combine and rearrange terms:
\[ \lim_{h \to 0}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc85e3b-dc2d-4f8b-a097-9d7c53c76aad%2F3a64ef8a-c770-44e9-a4d1-d94c51c1a180%2Fo3uc7qd_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:## Derivation of the Product and Quotient Rule
### Introduction
In calculus, we have learned that the derivative of the product and quotient of two differentiable functions can be computed using specific rules.
### The Product Rule
If \( f(x) \) and \( g(x) \) are two differentiable functions, the derivative of their product is given by:
\[ \frac{d}{dx} \left[ f(x)g(x) \right] = f(x)g'(x) + g(x)f'(x) \]
### The Quotient Rule
If \( f(x) \) and \( g(x) \) are two differentiable functions where \( g(x) \neq 0 \), the derivative of their quotient is:
\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2} \]
### Assignment 1: Derive the Product Rule
**Objective:** Derive the product rule formula using the definition of a derivative. Ensure every step is explained clearly for any Calculus student to follow.
#### Steps to Follow
1. **Definition of the Derivative:**
\[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
2. **Use a "magic trick":**
\[ -f(x + h)g(x) + f(x + h)g(x) \]
(This expression is 0, so essentially we did not change anything.)
3. **Factor out \( f(x + h) \) and \( g(x) \):**
\[ -f(x + h)g(x) + f(x + h)g(x) \]
To start you off, using the definition of a derivative:
\[ \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} \]
Now insert the "magic trick":
\[ \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h} \]
Combine and rearrange terms:
\[ \lim_{h \to 0}
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