Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Suppose \(y=f(x) g(x)\), where \(f(x)\) and \(g(x)\) are positive for all \(x\). Use logarithmic differentiation to find \(\frac{d y}{d x}\). Verify that your result is simply the Product Rule.

Step-by-Step Solution

Answer

Using logarithmic differentiation the derivative \(\frac{dy}{dx}\) is found to be \(f(x) g'(x) + g(x) f'(x)\), which is the same as the result achieved applying the Product Rule. Thus, the statement has been verified.

Step 1: Applying Logarithmic Differentiation

Start by taking the natural logarithm (ln) of both sides of the equation: \(ln(y) = ln[f(x) g(x)]\). By properties of logarithms, the right-hand side can be broken down as the sum of two separate logs which results to \(ln(y) = ln[f(x)] + ln[g(x)]\).

Step 2: Differentiating Both Sides

Derive both sides of the equation with respect to \(x\). The left-hand side becomes \(\frac{1}{y} \frac{dy}{dx}\) and the right-hand side becomes \(\frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)}\). This results in: \(\frac{1}{y} \frac{dy}{dx} = \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)}\)

Step 3: Solve for \(\frac{dy}{dx}\)

Multiplying both sides by \(y\), we get \(\frac{dy}{dx} = y[\frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)}]\). Substituting \(y = f(x)g(x)\) from the original equation, we finally arrive with the expression for \(\frac{dy}{dx} = f(x) g'(x) + g(x) f'(x)\)

Step 4: Verification through the Product Rule Application

The Product Rule for differentiation states that if \(y\) = uv, then the derivative \(dy/dx = u dv/dx + v du/dx\). Here, \(u = f(x)\) and \(v = g(x)\). Apply the rule and see that \(\frac{dy}{dx} = f(x) g'(x) + g(x) f'(x)\), which matches the step 3 result

Key Concepts

Product Rule

Understanding the product rule is essential for differentiating expressions where two functinos are multiplied together. Let's clarify what this rule is and how it works in the context of calculus.

The product rule states that if you have two differentiable functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) with respect to \( x \) is \( u'(x)v(x) + u(x)v'(x) \). In essence, you differentiate each function separately and sum the two possibilities—first, the derivative of the first function times the second function as is, and second, the first function as is times the derivative of the second function.

For instance, when applied to our exercise \( f(x)g(x) \), we differentiate \( f(x) \) and multiply it by \( g(x) \) and then differentiate \( g(x) \) and multiply it by \( f(x) \) and add the two products together for the result. This rule is powerful because it allows the differentiation of complex expressions without expanding them, making it a staple in calculus problem solving.

Natural Logarithm

The natural logarithm, denoted as \( ln(x) \), is a mathematical function that is the inverse of the exponential function \( e^x \). Essentially, if \( y = ln(x) \) then \( e^y = x \). The natural logarithm is unique since it uses the number \( e \) (approximately 2.71828), known as Euler's number, as its base. It's a fundamental constant in mathematics, particularly in calculus, and appears in many natural processes including growth and decay.

The properties of the natural logarithm, such as \( ln(ab) = ln(a) + ln(b) \), make it a handy tool in calculus—especially in logarithmic differentiation. In the context of our exercise, by taking the natural logarithm of the product of two functions, we can transform the product into a sum, simplifying the differentiation process since the derivative of a sum of functions is just the sum of their derivatives.

Derivative

The derivative is at the heart of calculus and is a measure of how a function changes as its input changes. Formally, it represents the slope of the tangent line to the function at a given point. The notation \( \frac{dy}{dx} \) reads as 'the derivative of \( y \) with respect to \( x \)'.

Derivatives can be understood as the instantaneous rate of change, which has applications in many fields, such as physics, where we might want to know the velocity (the rate of change of position) or economics, to understand the rate at which profit or cost changes with production. In our exercise, we calculate the derivative of the product of two functions using logarithmic differentiation, which relies on the fundamental property of derivatives of the natural logarithm and simplifies the process by breaking down products into sums.

Chain Rule

The chain rule is another crucial method in differential calculus and is used to calculate the derivative of a composite function. In its basic form, if \( y = g(f(x)) \) and we need \( \frac{dy}{dx} \) then the chain rule tells us to first find \( \frac{dg}{df} \) and \( \frac{df}{dx} \) and then multiply them together, \( \frac{dy}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} \).

The chain rule allows us to differentiate complex expressions by dividing the problem into simpler parts. It’s kind of like solving a puzzle by taking it piece by piece. In the context of logarithmic differentiation, we implicitly use the chain rule when differentiating the natural logarithm of a function, as in \( \frac{d}{dx}[ln(f(x))] \), which becomes \( \frac{f'(x)}{f(x)} \) by applying the chain rule.