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The product rule is a fundamental tool in calculus for finding the derivative of the product of two functions. If you have two functions, say \(u(x)\) and \(v(x)\) , and you need to find the derivative of their product, the product rule states that:
\[ (uv)' = u'v + uv' \]

This means you first take the derivative of \(u(x)\) , multiply it by \(v(x)\), and then add the product of \(u(x)\) and the derivative of \(v(x)\).

• Apply the product rule step-by-step to keep it simple.
• Make sure to differentiate each part clearly.

For example, in the given exercise, \( u = x^5 \) and \( v = \ln(3x) \) . Differentiating each separately as \( u' = 5x^4 \) and \( v' = \frac{1}{x} \) (using the chain rule which we will discuss later), we apply the product rule:

\[ g'(x) = (x^5)' \ln(3x) + x^5 (\ln(3x))' \]
This results in:

\[ g'(x) = 5x^4 \ln(3x) + x^5 \frac{1}{x} \]

Simplify further to:
\[ g'(x) = 5x^4 \ln(3x) + x^4 \]

Understanding and practicing the product rule with different functions help in mastering this concept.

The chain rule is another essential technique in calculus, especially when working with composite functions. If you have a composite function, say \( f(g(x)) \) , the chain rule helps differentiate it. The chain rule formula is:

\[ (f(g(x)))' = f'(g(x)) \cdot g'(x) \]

In simpler terms, you first differentiate the outer function, leaving the inner function untouched. Then, you multiply by the derivative of the inner function.

In our example, we deal with \( v = \ln(3x) \) . Here, \( f(u) = \ln(u) \) and \( u = 3x \) . Applying the chain rule, the derivative of the outer function \( \ln(u) \) is \( \frac{1}{u} \) , and multiplying by the derivative of the inner function \( 3x \) which is \( 3 \) , we get: \[ \frac{1}{3x} \cdot 3 = \frac{1}{x} \]

By practicing the chain rule, you will understand its importance in combining different rules of differentiation and correctly dealing with nested functions.

Understanding the natural logarithm function (\( \ln(x) \)) can simplify many differentiation problems. It is beneficial because it frequently appears in calculations, especially in problems involving growth and decay.

The natural logarithm, \( \ln(x) \) , is the inverse of the exponential function \( e^x \) . The key property for its differentiation is:

\[ (\ln(x))' = \frac{1}{x} \]

When the argument of the natural logarithm is more complex, like in \( \ln(3x) \) , you use the chain rule as discussed earlier. This gives:

\[ (\ln(3x))' = \frac{1}{3x} \cdot 3 = \frac{1}{x} \]


Remember, the natural logarithm is not defined for non-positive values of x. Be attentive to the domain of the function when dealing with \( \ln(x) \) . Here are some useful takeaways:

• Always check if the natural logarithm's argument is positive.
• When differentiating more complex natural logarithmic functions, apply the chain rule efficiently.
• Practice different derivatives involving \( \ln(x) \) to build a strong foundational understanding.

By mastering the differentiation of natural logarithms, you can easily tackle more advanced problems involving exponential and logarithmic functions.