Question

Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\)

Short answer

Question: Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f/g\) equals \(f^{\prime}/g^{\prime}\). Answer: One possible set of functions that satisfy this condition are \(f(x) = De^{k_2x}\) and \(g(x) = Ce^{k_1x}\), where \(C\), \(D\), \(k_1\), and \(k_2\) are constants.

Step-by-step solution

Finding the Derivative of a Quotient

To find the derivative of \(\frac{f}{g}\), we will use the quotient rule. The quotient rule states that if you have a function in the form \(h(x) = \frac{u(x)}{v(x)}\), then the derivative of \(h(x)\) with respect to \(x\) is: $$h^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{v(x)^2}$$ Applying the quotient rule to \(\frac{f}{g}\), we get: $$\frac{f}{g}^{\prime}(x) = \frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{g(x)^2}.$$

Set Up the Equation for the Required Condition

We want the derivative \(\frac{f}{g}^{\prime}\) to be equal to \(\frac{f^{\prime}}{g^{\prime}}\). So, we set up the equation as follows: $$\frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{g(x)^2}=\frac{f^{\prime}(x)}{g^{\prime}(x)}.$$

Cross-Multiplication

Let's cross-multiply to eliminate the fractions and manipulate the equation: $$f^{\prime}(x)g(x)g^{\prime}(x) - f(x)g^{\prime}(x)^2 = f^{\prime}(x)g(x)^2.$$

Identify Suitable Conditions for \(f\) and \(g\)

To make the equation true, we can either set both sides equal to zero or set each side equal to some function that will still satisfy the condition. Let us try setting both sides equal to zero: $$f^{\prime}(x)g(x)g^{\prime}(x) - f(x)g^{\prime}(x)^2 = 0 \Longrightarrow f^{\prime}(x)g(x)g^{\prime}(x) = f(x)g^{\prime}(x)^2.$$ Given that \(f\) and \(g\) are nonconstant functions, one possible way to satisfy the above equation is if: $$g^{\prime}(x) = k_1g(x) \quad \text{and} \quad f^{\prime}(x) = k_2f(x),$$ where \(k_1\) and \(k_2\) are constants.

Find Functions \(f\) and \(g\)

By solving the differential equations listed above, we can find \(f(x)\) and \(g(x)\): 1. Solve \(g^{\prime}(x) = k_1g(x)\): Integrating, we get \(g(x) = Ce^{k_1x}\), where \(C\) is a constant. 2. Solve \(f^{\prime}(x) = k_2f(x)\): Integrating, we get \(f(x) = De^{k_2x}\), where \(D\) is another constant. So, \(f(x) = De^{k_2x}\) and \(g(x) = Ce^{k_1x}\) are nonconstant functions that satisfy the given condition.

Key concepts

Derivative of a Quotient

When we deal with the derivative of a quotient, we encounter a scenario where one function is divided by another. Understanding the quotient rule is vital in calculus, especially when functions are not readily separable. The quotient rule can be expressed as:

\[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \]
Simply put, if you have a fraction where both the numerator and the denominator are functions of \(x\), their derivative isn't just the separate derivatives divided by each other. Instead, you subtract the product of the derivative of the numerator and the original denominator from the product of the original numerator and the derivative of the denominator, and then divide everything by the square of the original denominator. This rule ensures accuracy when working with complex rational functions, which are quite common in calculus problems.

Differential Equations

A differential equation is an equation that relates a function with its derivatives. In the context of the given exercise, differential equations are used to express possible forms of the functions \(f\) and \(g\) based on their rates of change. The equations

\[ g'(x) = k_1g(x) \]
and

\[ f'(x) = k_2f(x) \]
are simple first-order linear differential equations where the derivative of a function is proportional to the function itself. The solution to such equations is often an exponential function, as exponential functions are their own derivatives times a constant. This principle allows us to solve for the exact form of \(f(x)\) and \(g(x)\), which are essential in finding the original quotient under consideration.

Cross-Multiplication Method

The cross-multiplication method is a powerful tool for simplifying equations involving fractions. This method removes the denominators by multiplying each side of the equation by the denominators of the other side. In our problem, the goal was to set the derivative of a quotient equal to the quotient of the derivatives. To do this, we needed to perform cross-multiplication to obtain:

\[ f'(x)g(x)g'(x) - f(x)g'(x)^2 = f'(x)g(x)^2 \]
It transforms the equation into a simpler form without fractions, making it easier to manipulate and solve. Thus, cross-multiplication is not just an arithmetic trick; it's a crucial algebraic technique that facilitates the process of solving equations, especially when dealing with functions and their derivatives.

Nonconstant Functions

Nonconstant functions are the bedrock of calculus. They are functions whose value changes as the input changes, unlike a constant function that stays the same regardless of the input. The exercise explores nonconstant functions to demonstrate a practical understanding of calculus principles.

When given the differential equations \( g'(x) = k_1g(x) \) and \( f'(x) = k_2f(x) \), and knowing that the solution involves nonconstant functions, we deduce that the only functions that satisfy the condition of being proportional to their derivatives, through constants, are exponential functions. Therefore, the exercise not only highlights the purpose of the quotient rule but also illustrates how knowledge of differential equations and the nature of the functions involved lead to a comprehensive solution.