Problem 3 - Calculus Early Transcendentals

Question

Show that \(\frac{d}{d x}(\ln k x)=\frac{d}{d x}(\ln x),\) where \(x>0\) and \(k\) is a positive real number.

Step-by-Step Solution

Answer

Question: Show that \(\frac{d}{d x}(\ln k x)=\frac{d}{d x}(\ln x),\) where \(x>0\) and \(k\) is a positive real number. Answer: \(\frac{d}{d x}(\ln k x) = \frac{1}{x}\) and \(\frac{d}{d x}(\ln x) = \frac{1}{x}\), so the derivatives are equal. Therefore, \(\frac{d}{d x}(\ln k x)=\frac{d}{d x}(\ln x)\).

Step 1: Differentiate \(\ln(kx)\) with respect to \(x\)

To differentiate \(\ln(kx)\) with respect to \(x\), use the chain rule. The chain rule states that \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Let \(f(u) = \ln(u)\) and \(g(x) = kx\). Then, we have: \(f'(u) = \frac{1}{u}\) (this is the derivative of the natural logarithm) \(g'(x) = k\) (this is the derivative of a constant times a variable) Now, apply the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x) = \frac{1}{kx} \cdot k\) Simplify: \(\frac{d}{d x}(\ln k x) = \frac{k}{kx} = \frac{1}{x}\)

Step 2: Differentiate \(\ln(x)\) with respect to \(x\)

The derivative of \(\ln(x)\) with respect to \(x\) is: \(\frac{d}{d x}(\ln x) = \frac{1}{x}\)

Step 3: Compare the derivatives

Now compare the two derivatives. We have: \(\frac{d}{d x}(\ln k x) = \frac{1}{x}\) and \(\frac{d}{d x}(\ln x) = \frac{1}{x}\) Since the derivatives are equal, we have shown the required result: \(\frac{d}{d x}(\ln k x) = \frac{d}{d x}(\ln x)\)

Key Concepts

Chain Rule Calculus

Understanding the chain rule in calculus is vital for differentiating composite functions, which are functions composed of two or more functions. The chain rule can be stated as follows: If you have a function \(f(g(x))\), where \(f\) and \(g\) are both functions of \(x\), then the derivative of this composite function with respect to \(x\) is \(f'(g(x)) \times g'(x)\).

When differentiating a function like \(\ln(kx)\), where \(k\) is a constant, the chain rule allows us to differentiate the inner function \(g(x) = kx\) and the outer function \(f(u) = \ln(u)\) separately, and then multiply their derivatives to get the derivative of the whole composite function.

In simpler terms, we first differentiate \(\ln(u)\) concerning \(u\), which gives us \(\frac{1}{u}\), and then differentiate \(kx\) concerning \(x\), which gives us \(k\). These two results are then multiplied to give the derivative of \(\ln(kx)\) with respect to \(x\).

Natural Logarithm Derivative

The natural logarithm function, commonly denoted as \(\ln(x)\), is one whose derivative is crucial in calculus. The derived formula for the natural logarithm is quite simple: for any positive number \(x\), the derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\).

This rule stands as one of the fundamental building blocks when dealing with logarithmic differentiation. When you understand that the derivative of \(\ln(x)\) is \(\frac{1}{x}\), you can easily apply this knowledge to more complex logarithmic functions by using the chain rule for differentiation, as seen in the exercise above.

Implicit Differentiation

Implicit differentiation is a technique that is often applied when dealing with equations where the dependent variable, typically \(y\), cannot be easily isolated. In such cases, you assume that \(y\) is a function of \(x\) and differentiate both sides of the equation with respect to \(x\), applying the chain rule where necessary.

Even if this concept was not directly applied in the original exercise, it is important to understand, as implicitly differentiating functions is another method of finding the derivative without explicitly solving for \(y\). In the context of logarithmic differentiation, implicit differentiation can be particularly handy when dealing with more complex equations involving logarithms.

Calculus Proofs

Calculus proofs are rigorous arguments that confirm the validity of mathematical statements within calculus. Proofs may involve demonstrating that a certain differentiation strategy works, as in the textbook exercise example.

In the given exercise, the proof involved showing that \(\frac{d}{dx}(\ln(kx)) = \frac{d}{dx}(\ln(x))\), regardless of the value of constant \(k\). The proof required knowledge of derivative rules, including the chain rule, and an understanding of the natural logarithm derivative. By breaking down the proof into comprehensible steps, demonstrating each differentiation process, the result verifies the proposition with clarity and helps students understand the underpinning concepts more fully.