Problem 83 - Brief Applied Calculus

Question

Let \(f\) and \(g\) be differentiable functions of \(x\). Assume that denominators are not zero. Show that the Product Rule may be written in the following form: $$ \frac{d}{d x}(f \cdot g)=(f \cdot g)\left(\frac{f^{\prime}}{f}+\frac{g^{\prime}}{g}\right) $$

Step-by-Step Solution

Answer

The Product Rule can be rewritten as \((f \cdot g)\left(\frac{f'}{f} + \frac{g'}{g}\right)\).

Step 1: Write the Product Rule

The standard form of the Product Rule for differentiation states that for two differentiable functions \( f(x) \) and \( g(x) \), the derivative of their product is given by:\[ \frac{d}{dx}(f \cdot g) = f^{\prime} \cdot g + f \cdot g^{\prime} \]

Step 2: Express f' and g' in Terms of Ratios

Use the definition of a derivative to express \( f' \) and \( g' \) as \( \frac{f'}{f} \cdot f \) and \( \frac{g'}{g} \cdot g \) respectively, assuming the denominators are not zero. This gives us:\[ f' = \frac{f'}{f} \cdot f \] and \[ g' = \frac{g'}{g} \cdot g \]

Step 3: Substitute the Expressions

Replace \( f' \) and \( g' \) in the Product Rule expression with the expressions obtained in Step 2:\[ \frac{d}{dx}(f \cdot g) = \left( \frac{f'}{f} \cdot f \right) \cdot g + f \cdot \left( \frac{g'}{g} \cdot g \right) \]

Step 4: Factor Out the Product fg

Factor \( f \cdot g \) out of the resulting expression:\[ \frac{d}{dx}(f \cdot g) = f \cdot g \left(\frac{f'}{f} + \frac{g'}{g}\right) \] This shows that the Product Rule can be rewritten in the given form: \[ \frac{d}{d x}(f \cdot g) = (f \cdot g)\left(\frac{f^{\prime}}{f} + \frac{g^{\prime}}{g}\right) \]

Step 5: Verify the Form

Check if the rewritten Product Rule yields the same result as the original form:Given \( f \cdot g \left(\frac{f^{\prime}}{f} + \frac{g^{\prime}}{g}\right) \) indeed simplifies to \( f' \cdot g + f \cdot g' \) through distribution, confirming correctness.

Key Concepts

Differentiable Functions

A differentiable function is a function that has a derivative at each point in its domain. When we say a function is differentiable, it means we can find its derivative, which represents the slope of the tangent line at any point on the function's graph. Differentiability implies continuity, but the reverse is not necessarily true. If a function is differentiable on an interval, it is continuous there.

• The function must be smooth and free from any sharp corners or cusps in the interval in question.
• Calculus plays a crucial role in identifying and working with differentiable functions, providing insight into their properties and behavior.
• A function needs to be properly defined to be differentiable, without any interruptions in its curve.

Many real-world phenomena are modeled using differentiable functions because they provide a precise tool for analyzing how changing one quantity affects another. Understanding the concept of differentiability is fundamental in calculus as it lays the groundwork for studying motion, growth, and other dynamic processes.

Derivatives

Derivatives are central to calculus, representing the rate of change of a function concerning its variable. When we differentiate a function, we find its derivative, often noted as \(f'(x)\), \(f''(x)\), or \(\frac{df}{dx}\). The derivative tells us how the function's output value changes with respect to changes in the input value.

• In practical terms, the derivative can describe velocity in physics, indicating how quickly something moves over time.
• The process of finding a derivative is called differentiation, a fundamental operation in calculus.
• Partial derivatives and higher-order derivatives extend the utility of derivatives further, especially in multivariable calculus.

In our exercise, understanding derivatives is essential because they're used to reformulate the product rule in calculus. The product rule itself determines how we take derivatives of products of two functions by showing that \( (f\cdot g)' = f'\cdot g + f \cdot g' \). This rule is vital when dealing with products of differentiable functions, helping to dissect how combined functions change.

Factorization in Calculus

Factorization in calculus isn't too different in concept from algebraic factorization. It involves rewriting an expression as a product of its factors. In the context of derivatives and the product rule, factorization simplifies expressions and calculations.

• In the exercise, we factor the expression \(f \cdot g\) from the product rule rewritten form: \((f\cdot g)' = (f \cdot g)\left(\frac{f'}{f} + \frac{g'}{g}\right)\).
• Factorization allows us to express complex derivatives in a more manageable form, aiding in their understanding and application.
• Through factorization, we reveal underlying relationships and simplify computational work, especially important in lengthy or convoluted expressions.

Factorization isn't just about making equations look cleaner; it aids in mathematical reasoning, allowing for further manipulation or integration of expressions. By simplifying complex derivatives into products of simpler terms, we make the calculus of these functions much more accessible.