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The product rule is a fundamental principle in calculus used to differentiate functions that are the product of two other functions. When you have a function that looks like \( f(x) = u(x) \cdot v(x) \), the product rule comes into play. This rule is essential because not all functions are simply sums or differences where you can apply basic rules of differentiation directly.

The formula for the product rule is:

• \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

Let's break it down:

• \( u'(x) \cdot v(x) \): Differentiate the first function \( u(x) \), then multiply by the second function \( v(x) \).
• \( u(x) \cdot v'(x) \): Leave the first function \( u(x) \) as is, then multiply by the derivative of the second function \( v(x) \).

Using the product rule ensures that we correctly account for both functions' rates of change when they are multiplied.

The chain rule is another key tool in differentiation, used when dealing with composite functions. Composite functions are functions within functions, which often appear when you need to differentiate more complex expressions.

For a function that can be expressed as \( h(x) = g(f(x)) \), the chain rule allows us to find its derivative. Its formula is:

• \( h'(x) = g'(f(x)) \cdot f'(x) \)

Here's how it works:

• Differentiate the outer function \( g(x) \) as if it's alone, but keep the inner function \( f(x) \) inside.
• Then multiply by the derivative of the inner function \( f(x) \).

In the exercise, the chain rule is applied to \( \ln(5x) \). First, you differentiate \( \ln(5x) \) to get \( \frac{1}{5x} \). Then, you multiply by the derivative of \( 5x \), which is 5, resulting in \( \frac{1}{x} \). This specific step of using the chain rule is vital whenever your differentiation involves nested functions.

Trigonometric functions, such as sine and cosine, are pervasive in both mathematics and the real world, often depicting wave-like phenomena or periodic behavior.

When differentiating trigonometric functions, there are some rules to remember:

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

In our exercise, we use the derivative of \( \sin x \). The differentiation directly gives us \( \cos x \).

Understanding the differentiation of trigonometric functions is crucial since these derivatives frequently appear in physics, engineering, and various applied sciences. Their behavior often influences how wave forms and oscillatory motions are analyzed and interpreted.

Logarithmic functions are inverses of exponential functions and have unique properties that make them essential in various applications, including growth and decay modeling, and simplifying multiplication into addition.

When differentiating simple logarithmic functions, such as \( \ln(x) \), the derivative is \( \frac{1}{x} \). This property is pivotal in calculus, making logarithmic functions easier to manage algebraically.

In our specific exercise, we differentiate \( \ln(5x) \) using the chain rule. The general approach here is:

• First take the derivative of \( \ln(u) \) as \( \frac{1}{u} \). Here, \( u = 5x \).
• Then apply the chain rule by multiplying that result with the derivative of \( 5x \), which is 5, leading to \( \frac{1}{x} \).

Logarithmic derivatives often simplify complex multiplication and division operations into simpler additive forms, vastly simplifying calculus tasks as shown in this example.