91Ó°ÊÓ

The product rule is an essential tool in calculus for differentiating products of two functions. If we have a function that is the product of two other functions, say, \(u(x)\) and \(v(x)\), then the derivative of their product is given by:

• \((uv)' = u'v + uv'\)

This means you take the derivative of the first function and multiply it by the second function, and then add it to the first function multiplied by the derivative of the second function.
In the exercise, the function \(3x \cdot \ln(x)\) appears during differentiation. Here, \(3x\) and \(\ln(x)\) are treated as separate functions. To differentiate \(3x \cdot \ln(x)\), apply the product rule:

• The derivative of \(3x\) is 3.
• \(\ln(x)\) remains the same.
• The derivative of \(\ln(x)\) is \(\frac{1}{x}\).

Using these, the application of the product rule yields \(3 \ln(x) + 3 \cdot \frac{1}{x} \cdot x\), simplifying to \(3 \ln(x) + 3\).

The chain rule is a fundamental principle in calculus used when differentiating composite functions. A composite function is a function wrapped inside another function, generally expressed as \(y = g(f(x))\).
The chain rule formula is:

• \(\frac{dy}{dx} = \frac{dg}{df} \times \frac{df}{dx}\)

In simple terms, differentiate the outer function, and then multiply it by the derivative of the inner function.
In our step-by-step solution, the chain rule is applied when differentiating the left side: \(\frac{d}{dx}[\ln(f(x))]\). Here, the outer function is \(\ln(u)\), and the inner function is \(u = f(x)\).
The derivative of \(\ln(u)\) is \(\frac{1}{u}\). Hence:

• \(\frac{d}{dx}[\ln(f(x))] = \frac{1}{f(x)} \cdot f'(x)\)

This chain rule application is crucial for isolating \(f'(x)\) in the solution.

The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is approximately 2.718, a fundamental constant in mathematics.
The natural logarithm has some unique properties which are particularly useful in calculus:

• \(\ln(ab) = \ln(a) + \ln(b)\)
• \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• \(\ln(a^b) = b\ln(a)\)

In our exercise, a critical step involves applying \(\ln(x^{3x})\). Using the property \(\ln(a^b) = b\ln(a)\), it simplifies to \(3x \cdot \ln(x)\). This simplification is key to making differentiation more straightforward, and it is essential to understand these properties when dealing with logarithmic differentiation.
This technique helps transform complex expressions into simpler ones, which are often easier to differentiate.