Problem 75 - Calculus and its applications

Question

A proof of the Product Rule appears below. Provide a justification for each step. a) \(\frac{d}{d x}[f(x) \cdot g(x)]=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x) g(x)}{h}\) b) \(\quad=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x+h) g(x)+f(x+h) g(x)-f(x) g(x)}{h}\) c) \(\quad=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x+h) g(x)}{h}+\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)}{h}\) d) \(\quad=\lim _{h \rightarrow 0}\left[f(x+h) \cdot \frac{g(x+h)-g(x)}{h}\right]+\lim _{h \rightarrow 0}\left[g(x) \cdot \frac{f(x+h)-f(x)}{h}\right]\) e) \(\quad=f(x) \cdot \lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}+g(x) \cdot \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) f) \(\quad=f(x) \cdot g^{\prime}(x)+g(x) \cdot f^{\prime}(x)\) g) \(\quad=f(x) \cdot\left[\frac{d}{d x} g(x)\right]+g(x) \cdot\left[\frac{d}{d x} f(x)\right]\)

Step-by-Step Solution

Answer

The proof shows the Product Rule: \( \frac{d}{dx}[f(x)g(x)] = f(x)g'(x) + g(x)f'(x) \).

Step 1: Definition of the Derivative

The expression given in part (a) uses the definition of a derivative for the product of two functions \( f(x) \) and \( g(x) \). It is a general restatement of the derivative as the limit of the difference quotient as \( h \) approaches zero.

Step 2: Adding and Subtracting the Same Term

In part (b), we add and subtract the term \( f(x+h)g(x) \) which helps in simplifying the expression into forms that allow for separation according to each function \( f \) and \( g \). This technique often helps in proof by creating a difference that can be split.

Step 3: Splitting the Limit

In part (c), the expression from part (b) is split into two separate limits. This is detailed as follows: the first limit concerning \( g \) holding \( f(x+h) \) constant, and the second limit concerning \( f \) holding \( g(x) \) constant. This showcases each product term independently.

Step 4: Factoring Within Limits

For part (d), factor \( f(x+h) \) and \( g(x) \) out of their respective limits. This exploits the distributive property of limits, separating constants from limits. Each of these sections now resembles the basic derivative form (difference quotient).

Step 5: Applying the Definition of Derivatives

In part (e), we apply the definition of the derivative to\( \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \) which equals \( g'(x) \), and \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \) which equals \( f'(x) \).

Step 6: Final Algebraic Simplification

Part (f) states the final simplified expression based on our transformation: \( f(x) \cdot g'(x) + g(x) \cdot f'(x) \), which represents the derivative of a product of two functions - the Product Rule.

Step 7: Equating to Product Rule Notation

In part (g), the derivative expressions are rewritten with derivative notation \( \frac{d}{dx}g(x) \) and \( \frac{d}{dx}f(x) \), reinforcing that this represents the formal expression of the Product Rule in calculus.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus. It measures how a function changes as its input changes. Essentially, the derivative of a function at a point tells us the rate at which the function is changing at that particular point. It is often denoted as \( f'(x) \) or \( \frac{d}{dx}f(x) \). Calculating a derivative involves finding the limit of the difference quotient as the change in the input \( h \) approaches zero.

• The notation \( \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \) captures the essence of how a derivative works.
• It's like asking, "If you make a tiny change in \( x \), how much does \( f(x) \) change?"

Understanding derivatives is crucial because they provide information about the slope of a curve at any given point. They are extensively used in various applications like physics, engineering, and even economics.

Limit

The concept of a limit is central to calculus and particularly fundamental when discussing derivatives. A limit helps us understand the behavior of a function as its input gets infinitely close to some value.

• In derivative calculations, limits allow us to rigorously define how we can approximate instantaneous rates of change.
• The expression \( \lim_{h \to 0} \) is crucial, as it is used to find the derivative by focusing on what happens as the increment \( h \) becomes very small.

Thus, the limit is the foundation upon which the concept of a derivative stands, making it possible to deal with and understand continuous change effectively.

Difference Quotient

The difference quotient is a formula used to calculate the derivative and is integral to proving the Product Rule. It represents the slope of the secant line between two points on the graph of a function.

• The standard form of a difference quotient is \( \frac{f(x+h)-f(x)}{h} \).
• It measures the average rate of change of the function over an interval \( h \).
• The difference quotient is a stepping stone to finding the derivative as \( h \) approaches zero.

Breaking apart the difference quotient allows proofs, like the Product Rule, to separate components and simplify the expression through algebraic manipulation, making complex derivatives easier to derive.

Calculus Proving Techniques

Calculus proofs often involve clever techniques such as algebraic manipulation, strategic splitting, and factorization to show how rules work behind the scenes. The Product Rule proof shows these techniques clearly.

• Strategically adding and subtracting similar terms, as seen in step (b), is common to transform complex expressions.
• Splitting into separate limits, like in step (c), lets us handle each part of a function independently.
• The use of factorization allows limits to simplify effectively, as shown in step (d).

These proving techniques help generalize important rules, enabling us to apply them broadly across varied mathematical problems, reinforcing a deep understanding of calculus principles.