Problem 31 - University Calculus: Early Transcendentals

Question

Find \(d p / d q\). $$p=\frac{q \sin q}{q^{2}-1}$$

Step-by-Step Solution

Answer

Use the quotient rule on \(p=\frac{q \sin q}{q^2 - 1}\) to find the derivative.

Step 1: Identify the Function

We have the function \(p = \frac{q \sin q}{q^2 - 1}\). We need to find the derivative \(\frac{dp}{dq}\). This is a quotient of two functions of \(q\).

Step 2: Apply the Quotient Rule

The quotient rule is \(\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\). Here, let \(u = q \sin q\) and \(v = q^2 - 1\).

Step 3: Find Derivative of the Numerator (u)

For \(u = q \sin q\), apply the product rule: \(\frac{du}{dq} = \sin q + q \cos q\).

Step 4: Find Derivative of the Denominator (v)

For \(v = q^2 - 1\), the derivative is \(\frac{dv}{dq} = 2q\).

Step 5: Substitute into the Quotient Rule

Substitute \(u\), \(\frac{du}{dq}\), \(v\), and \(\frac{dv}{dq}\) into the quotient rule. The derivative \(\frac{dp}{dq} = \frac{(q^2-1)(\sin q + q\cos q) - (q\sin q)(2q)}{(q^2-1)^2}\).

Step 6: Simplify the Expression

Simplify the expression for \(\frac{dp}{dq}\). This involves expanding the terms and combining like terms, but since the problem does not require simplification, we leave it at: \[ \frac{dp}{dq} = \frac{(q^2-1)(\sin q + q\cos q) - 2q^2 \sin q}{(q^2-1)^2} \]

Key Concepts

Derivative

A derivative represents the rate at which a function is changing at any given point and is a fundamental concept in calculus. For a function expressed as \(p(q)\), finding \(\frac{dp}{dq}\) means determining how \(p\) changes with respect to small changes in \(q\). In this specific exercise, the function \(p\) is expressed as a fraction of two functions of \(q\), specifically \(p = \frac{q \sin(q)}{q^2 - 1}\). Here, identifying the derivative helps us understand the behavior of this function over its domain. When working with such expressions, utilizing the appropriate rules for differentiation, such as the product and quotient rule, becomes necessary. This ensures accuracy in the process of finding how each part of the function contributes to changes in \(p\).

Understanding the derivative is not just about calculating it but also comprehending what it represents—a snapshot of the function's behavior at any particular \(q\) value.

Product Rule

The product rule is essential when taking the derivative of products of functions, like \(u(q) = q \sin(q)\) in our exercise. The product rule states that if you have two differentiable functions, \(u\) and \(v\), their product's derivative is given by: \[ \frac{d}{dq}(u \cdot v) = u'(q)v(q) + u(q)v'(q) \]In this example, with \(u = q\) and \(v = \sin(q)\), the product rule helps us find the derivative of \(u\):

• The derivative of \(q\) is 1.
• The derivative of \(\sin(q)\) is \(\cos(q)\).

Applying the product rule, we derive \(\frac{du}{dq} = \sin(q) + q \cos(q)\). Recognizing when and how to use the product rule is key.

It allows one to break down complex differentiations into manageable parts, making the math more straightforward.

Simplification

Simplification is the process of reducing an expression to its most basic form, making it easier to interpret and use. After applying rules like the quotient and product rule, you'll often end up with a somewhat cluttered expression. This particular example ends with:\[ \frac{dp}{dq} = \frac{(q^2-1)(\sin(q) + q\cos(q)) - 2q^2 \sin(q)}{(q^2-1)^2} \]While simplifying isn't always mandatory, it can make complex expressions more intuitive or reveal certain behaviors or symmetries. The simplification process involves combining like terms and clearing any redundancies or unnecessary complexity.

In mathematics, simplification doesn't change the underlying relationship of an equation but makes it easier to understand or integrate into further calculations. Even if the example doesn't require further simplification, it is a crucial final step in many problems to make the results more accessible, especially in applications or analysis.

Function Identification

Before applying derivative rules, identifying the components of the function is crucial. In this exercise, we start by acknowledging that we have a quotient situation, \(p = \frac{q \sin(q)}{q^2 - 1}\), which calls for the quotient rule. In this function:

• The numerator \(u\) is \(q \sin(q)\).
• The denominator \(v\) is \(q^2 - 1\).

Identifying these correctly ensures the subsequent steps of applying derivative rules are correct. Each part of the fraction has its own structure, requiring its own focus and treatment in the differentiation process.

By clearly identifying the necessary functions and recognizing which rule applies, we set a firm foundation for accurate derivative calculation. This step might seem simple, but it is vital in avoiding errors in multi-step mathematical problems.