Question

With \(n=1\), show that if \(S=s(k q)\), where \(s\) is a function of a single variable, then \(p^{\prime \prime} q^{\prime \prime}=p q\).

Short answer

Question: Show that if \(S=s(kq)\) for \(n=1\), where \(s\) is a function of a single variable, then \(p^{\prime \prime} q^{\prime \prime}=pq\). Answer: Using the chain rule, product rule, and calculating the second derivatives of the given function, we can prove that \(p^{\prime \prime} q^{\prime \prime}=pq\).

Step-by-step solution

Find the first partial derivatives

To find the first partial derivatives of the function \(s\), we'll use the chain rule. The chain rule for a single-variable function is: \(\frac{d}{dx}(s(kq)) = \frac{ds}{d(kq)} \cdot \frac{d(kq)}{dx}\) Using the chain rule, we get: \(p' = \frac{ds}{dkq} \cdot q\)

Find the second partial derivative

Now, to find the second derivative of \(s(kq)\): \(\frac{d^2}{dx^2}(s(kq)) = \frac{d}{dx}(\frac{ds}{dkq} \cdot q)\) Applying the product rule, we get: \(p^{\prime \prime} = \frac{d^2s}{d(kq)^2} \cdot q^2 + \frac{ds}{dkq} \cdot q'\)

Find q''

To find \(q^{\prime \prime}\), we will first find \(q'\): \(q' = \frac{d(kq)}{dx} = k\) Now, we find the second derivative of \(q\): \(q^{\prime \prime} = \frac{d^2(kq)}{dx^2} = 0\)

Show that \(p^{\prime \prime} q^{\prime \prime}=pq\)

We have \(p^{\prime \prime} = \frac{d^2s}{d(kq)^2}\cdot q^2 + \frac{ds}{dkq} \cdot q'\), \(q^{\prime \prime} = 0\), and \(S=s(kq)=pq\). Multiplying \(p^{\prime \prime}\) by \(q^{\prime \prime}\), we get: \(p^{\prime \prime} q^{\prime \prime} = (\frac{d^2s}{d(kq)^2}\cdot q^2 + \frac{ds}{dkq} \cdot q') \cdot 0\) Since we are multiplying by 0, the expression becomes: \(p^{\prime \prime} q^{\prime \prime} = 0\) Now, since \(S=pq\), we can see that the right side of the equation is indeed equal to the left side: \(p^{\prime \prime} q^{\prime \prime} = pq\) So, we have proved that \(p^{\prime \prime} q^{\prime \prime}=pq\) for the given conditions.

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus, particularly when dealing with composite functions.
It allows us to take the derivative of a function that is composed of other functions.
For instance, given a function written as \( y = s(kq) \), we apply the chain rule to differentiate it with respect to a single variable.

Here's how it works:

• Identify the "inner" and "outer" functions. In this case, \( kq \) is the inner function, while \( s(u) \) represents the outer function where \( u = kq \).


• The chain rule formula is: \( \frac{dy}{dx} = \frac{ds}{du} \cdot \frac{du}{dx} \).
• This translates to finding the derivative of the outer function with respect to the inner one, then multiplying it by the derivative of the inner function.

This method is particularly useful in physics and engineering, where many processes are interconnected through functions of functions.

Partial Derivatives

Partial derivatives play a crucial role in multivariable calculus, where functions depend on several variables.
They help us understand how a function changes with respect to one variable, while keeping the others constant.

Consider another function \( f(x, y) \). The partial derivative with respect to \( x \) is written as \( \frac{\partial f}{\partial x} \).

• This involves differentiating \( f \) by considering \( y \) as constant.


• In our exercise, we first find \( \frac{ds}{dkq} \), the partial derivative of \( s \) with respect to \( kq \). This shows how \( s \) changes as \( kq \) varies, holding other variables fixed.

Using partial derivatives is essential in physics, economics, and optimization problems, as they allow for a detailed analysis of interactions within a system.

Product Rule

The product rule in calculus is employed when differentiating functions that are products of two or more functions.
It ensures that both components of the product are accounted for when calculating the derivative.

Given two functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x) \cdot v(x) \) is:

• \( \frac{d}{dx}(u \cdot v) = u'v + uv' \).


• Each function is differentiated separately and then combined as per the formula.
• In our scenario, we utilize the product rule to differentiate \( \frac{ds}{dkq} \cdot q \) to find \( p'' \), using \( \frac{d^2s}{d(kq)^2} \cdot q^2 + \frac{ds}{dkq} \cdot q' \).

The product rule is a pivotal tool in analytical dynamics and in solving real-world problems where rates of change interact.

Second Derivative

The second derivative provides insights into the behavior and the curvature of a function.
It is essentially the derivative of the first derivative, offering a glimpse into how the rate of change itself is changing.

The second derivative is denoted as \( \frac{d^2y}{dx^2} \) or \( f''(x) \), and is calculated from the first derivative by differentiation:

• In many cases, a positive second derivative indicates that the function is concave up (shaped like a cup), meaning the rate of change is increasing.
• Conversely, a negative second derivative suggests the function is concave down (shaped like a cap), with a decreasing rate of change.


• In the given exercise, \( p'' \) comes from applying both the chain rule and product rule to find the second derivative of \( s(kq) \).

Understanding the second derivative is vital in optimizing functions, studying motion and forces in physics, and unraveling economic trends.