Problem 74 - University Calculus: Early Transcendentals

Question

The Reciprocal Rule a. The Reciprocal Rule says that at any point where the function \(v(x)\) is differentiable and different from zero, $$\frac{d}{d x}\left(\frac{1}{v}\right)=-\frac{1}{v^{2}} \frac{d v}{d x}$$ Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule.

Step-by-Step Solution

Answer

a) By setting \( u = 1 \), the Quotient Rule simplifies to the Reciprocal Rule. b) Expressing the derivative of \( \frac{u}{v} \) as a product shows how it integrates both rules.

Step 1: Applying the Quotient Rule to Part (a)

The Quotient Rule states that the derivative of a quotient \( \frac{u}{v} \) is given by \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] To solve part (a), set \( u = 1 \), then \( \frac{du}{dx} = 0 \). The Quotient Rule becomes \[ \frac{d}{dx} \left( \frac{1}{v} \right) = \frac{v \cdot 0 - 1 \cdot \frac{dv}{dx}}{v^2} = -\frac{1}{v^2} \frac{dv}{dx} \] which matches the Reciprocal Rule formula.

Step 2: Expressing Part (b) with the Quotient Rule Using the Product Rule

The Product Rule states \( \frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx} \). For a function \( \frac{u}{v} \), express it as \( u \cdot \frac{1}{v} \). Then apply the Product Rule: \[ \frac{d}{dx} \left( u \cdot \frac{1}{v} \right) = \frac{du}{dx} \cdot \frac{1}{v} + u \cdot \frac{d}{dx} \left( \frac{1}{v} \right) \] Substituting the Reciprocal Rule derivative, this becomes: \[ \frac{du}{dx} \cdot \frac{1}{v} - u \cdot \frac{1}{v^2} \cdot \frac{dv}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] Thus, showing how the Reciprocal Rule with Product Rule yields the Quotient Rule.

Key Concepts

Reciprocal Rule

The Reciprocal Rule is a handy tool in calculus for finding derivatives of reciprocal functions. This rule states that if you have a function of the form \( \frac{1}{v} \), where \( v(x) \) is differentiable and does not equal zero, the derivative is given by:

• \( \frac{d}{dx}\left(\frac{1}{v}\right) = -\frac{1}{v^2} \frac{dv}{dx} \).

This formula can be derived using the Quotient Rule by setting the numerator \( u = 1 \), which simplifies the process. When \( u \) is constant, its derivative \( \frac{du}{dx} = 0 \), making the formula fit perfectly into the Reciprocal Rule form.
The Reciprocal Rule underscores how a complex-looking derivative can be simplified using known rules, thereby simplifying the overall problem-solving process.

Quotient Rule

The Quotient Rule is used to differentiate functions that are ratios of two differentiable functions, \( u(x) \) and \( v(x) \). The formula for the Quotient Rule is:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).

This rule breaks down the differentiation process into manageable components by operating on the numerator and the denominator separately.
To relate it back to the Reciprocal Rule, if \( u = 1 \), the Quotient Rule simplifies to give the Reciprocal Rule's formula. This highlights the Quotient Rule's flexibility in accommodating special cases, such as differentiating the reciprocal of a function.
Through practice, understanding and applying the Quotient Rule becomes more intuitive, serving as an essential tool in calculus for managing complex expressions.

Product Rule

The Product Rule is essential for differentiating products of two functions. If you have two functions, \( u(x) \) and \( v(x) \), their derivative is given by:

• \( \frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx} \).

This rule tells us that to differentiate a product, you take the derivative of the first function multiplied by the second, and add the first function multiplied by the derivative of the second.
In the context of proving the Quotient Rule from the Reciprocal Rule, the Product Rule plays a critical role. By expressing a quotient \( \frac{u}{v} \) as a product \( u \cdot \frac{1}{v} \) and applying the Product Rule, alongside the Reciprocal Rule, we can derive the Quotient Rule. This demonstrates the connectivity and versatility of breakdown methods in calculus.

Differentiation

Differentiation is the fundamental operation in calculus that finds the rate at which a function is changing. Whether using the Reciprocal, Quotient, or Product Rule, the goal is to find the derivative, which is essentially the slope of the tangent line at any point on a function.
Various rules, like the Reciprocal, Quotient, and Product Rules, provide shortcuts and methods to simplify the differentiation of complex functions. They enable you to tackle derivatives in a structured way, ensuring you capture the changing rate accurately.

• The Reciprocal Rule helps with reciprocal functions.
• The Quotient Rule assists with functions presented as fractions.
• The Product Rule simplifies the differentiation of products of two functions.

Understanding these concepts helps build a strong foundation in differential calculus, allowing you to explore more intricate mathematical models and real-world applications with confidence.