Problem 14 - Calculus: Early Transcendental Functions

Question

Evaluate the derivative using properties of logarithms where needed. $$\frac{d}{d x}\left[\ln \left(x^{5} \sin x \cos x\right)\right]$$

Step-by-Step Solution

Answer

The derivative of the function \(\ln \left(x^{5} \sin x \cos x\right)\) is \(\frac{5}{x} + \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}\).

Step 1: Simplify using the properties of logarithms

Using the rule \(\ln ab = \ln a + \ln b\), the logarithm can be expanded and simplified to: \(\ln (x^{5} \sin x \cos x)= \ln(x^{5})+\ln(\sin x)+\ln(\cos x)\). Then \(\ln(x^{5})\) can be rewritten as \(5\ln(x)\) according to the rule \(\ln(a^n) = n\ln a\). So the final equation will be: \(5\ln(x)+\ln(\sin x)+\ln(\cos x)\).

Step 2: Differentiate each term separately

Next, differentiate each term in the simplified equation. The derivative of \(\ln(x)\) is \(\frac{1}{x}\), so the derivative of \(5\ln(x)\) will be \(5 \cdot \frac{1}{x}\). The derivative of \(\ln(\sin x)\) is \(\frac{\cos x}{\sin x}\) and the derivative of \(\ln(\cos x)\) is \(\frac{-\sin x}{\cos x}\).

Step 3: Summarize and simplify the differentiated function

Adding all the differentiated terms give the final differentiated function: \(\frac{5}{x} + \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}\). It gives the derivative of the original function.

Key Concepts

Properties of Logarithms

Logarithms have a set of properties that are extremely useful in simplifying complex expressions. One of the most important properties is that the logarithm of a product is the sum of the logarithms of its factors:

• \( \ln(ab) = \ln(a) + \ln(b) \)
• \( \ln(a^n) = n\ln(a) \)

These properties let us break down complex logarithmic expressions into simpler parts which can be tackled individually. In the exercise, we used these principles to simplify \( \ln(x^5 \sin x \cos x) \) into \( 5\ln(x) + \ln(\sin x) + \ln(\cos x) \). This made it much easier to take the derivative. Remembering and using these properties will help you in not just solving this kind of problem, but many others where logarithmic functions are involved.
Understanding how to apply these properties is critical because it allows transformations that simplify the differentiation process.

Differentiation Rules

Differentiation is a fundamental concept in calculus which involves finding the rate at which a function changes at any given point. There are specific rules that help in differentiating functions, known as differentiation rules. Some commonly used ones include:

• Power Rule: For \( x^n \), the derivative is \( nx^{n-1} \)
• Constant Rule: Derivative of a constant is 0
• Sum Rule: Derivative of a sum of functions is the sum of the derivatives
• Chain Rule: Used for composite functions, which we'll discuss in more detail later

In our exercise, we specifically used the derivatives of logarithmic functions such as \( \ln(x) \), where the derivative is \( \frac{1}{x} \). Differentiation rules simplify our jobs by providing a structured method for calculating derivatives, ensuring even complex expressions are solvable.

Trigonometric Functions

Trigonometric functions such as \( \sin(x) \) and \( \cos(x) \) often appear in calculus problems, and they have specific derivatives that are crucial to know:

• The derivative of \( \sin(x) \) is \( \cos(x) \)
• The derivative of \( \cos(x) \) is \( -\sin(x) \)

Understanding these basic derivatives allows us to tackle more complicated expressions, especially when combined with logarithmic functions as in the original exercise. When differentiating \( \ln(\sin x) \) and \( \ln(\cos x) \), these derivatives become parts of the final answer. Such trigonometric derivatives are an essential part of calculus, appearing frequently in application problems like wave equations and oscillations.

Chain Rule

The Chain Rule is a powerful technique used for differentiating composite functions. A composite function is one that involves a function inside another function, like \( f(g(x)) \). The Chain Rule states that:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In the exercise, we differentiate \( \ln(\sin x) \) which essentially translates to applying the chain rule. We treat \( \ln u \) and set \( u = \sin x \); then it's differentiated as \( \frac{1}{u} \cdot (u') \). Hence, it becomes \( \frac{cos x}{sin x} \). The Chain Rule is essential for dealing with layered functions, ensuring that multi-step differentiation is made straightforward and systematic.