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Logarithmic functions are inverse operations to exponential functions. They help us solve equations where the unknown appears as the exponent. In their simplest form, a logarithm answers the question, 'To what exponent must a given base be raised, to produce a specific number?' For example, the logarithm base 10 of 100 is 2, because 10 raised to the power 2 is 100. Logarithms have different notation based on their base:

• If the base is 10, it is common logarithm, written as \( \log_{10}(x) \).
• If the base is \( e \) (approximately 2.718), it is the natural logarithm, written as \( \ln(x) \).

In the differential calculus context, when we see \( \log(x) \) without explicit mention of a base, it usually refers to the natural logarithm \( \ln(x) \). This is why in the exercise, \( G(x)=\log(5x+4) \) is treated as \( \ln(5x+4) \) for differentiation purposes.

When you need to differentiate a composite function, the chain rule becomes an indispensable tool. A composite function is a function nested inside another function.
For instance, in the expression \( G(x)=\ln(5x+4) \), \( 5x+4 \) is the inner function, and \( \ln(u) \) is the outer function where \( u=5x+4 \).The chain rule states that to find the derivative of \( f(g(x)) \), you take:
1. The derivative of the outer function \( f \) with respect to the inner function \( u \), and
2. Multiply it by the derivative of the inner function \( g \) with respect to \( x \).Mathematically, it is expressed as:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]This approach streamlines the process of differentiation, especially with functions that seem complex at first glance.

Calculating derivatives allows us to understand how a function changes at any given point. It involves finding the rate at which one quantity changes with respect to another. Here, the focus is on differentiating logarithmic functions using the chain rule.
Let's take \( G(x)=\ln(5x+4) \) as an example:

• **Differentiate the Outer Function:** The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). Thus, if \( u = 5x+4 \), the derivative is \( \frac{1}{5x+4} \).
• **Differentiate the Inner Function:** The inner function is a simple linear function \( 5x + 4 \). Its derivative with respect to \( x \) is 5.

Putting it all together using the chain rule, we multiply these derivatives to find: \[G'(x) = \frac{1}{5x+4} \cdot 5 = \frac{5}{5x+4}\]This formula gives us the derivative \( G'(x) \), showing how \( G(x) \) changes as \( x \) changes.