Problem 90 - Calculus and its applications

Question

Differentiate. $$ g(x)=\ln (2 x) \cdot \ln (7 x) $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{1}{x} \ln(14x^2) \).

Step 1: Recognize the Product Rule

To differentiate the function \( g(x) = \ln(2x) \cdot \ln(7x) \), notice that it is a product of two functions \( u(x) = \ln(2x) \) and \( v(x) = \ln(7x) \). We'll need to use the product rule which states \((uv)' = u'v + uv'\).

Step 2: Differentiate \( u(x) = \ln(2x) \)

To find \( u'(x) \), use the chain rule on \( \ln(2x) \). The derivative of \( \ln(x) \) is \( \frac{1}{x} \), and the derivative of \( 2x \) is \( 2 \). Thus, \( u'(x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \).

Step 3: Differentiate \( v(x) = \ln(7x) \)

Similarly, apply the chain rule to differentiate \( v(x) = \ln(7x) \). The derivative of \( \ln(x) \) is \( \frac{1}{x} \), and the derivative of \( 7x \) is \( 7 \). Therefore, \( v'(x) = \frac{1}{7x} \cdot 7 = \frac{1}{x} \).

Step 4: Apply the Product Rule

Using the product rule, substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into \((uv)' = u'v + uv'\). This gives: \[ g'(x) = \left(\frac{1}{x}\right) \ln(7x) + \ln(2x) \left(\frac{1}{x}\right). \]

Step 5: Simplify the Expression

Factor out \( \frac{1}{x} \) from both terms in the expression: \[ g'(x) = \frac{1}{x} \big( \ln(7x) + \ln(2x) \big). \] Use the logarithmic identity \( \ln(a) + \ln(b) = \ln(ab) \) to further simplify: \[ g'(x) = \frac{1}{x} \ln(14x^2). \]

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus for differentiating products of two functions. When you encounter an expression that involves two functions multiplied together, such as \( g(x) = u(x) \cdot v(x) \), the product rule is your best friend. This rule states:

• \((uv)' = u'v + uv' \)

In simple terms, this means that to find the derivative of the product of two functions, differentiate the first function \( u \) and multiply it by the second function \( v \), then add the product of the first function and the derivative of the second function.
For example, in our exercise, to differentiate \( g(x) = \ln(2x) \cdot \ln(7x) \), we treat \( \ln(2x) \) as \( u(x) \) and \( \ln(7x) \) as \( v(x) \). Using the product rule, we find that the derivative of \( g(x) \) is:

• \( g'(x) = u'(x)v(x) + u(x)v'(x) \)
• Or: \( \left( \frac{1}{x} \right) \ln(7x) + \ln(2x) \left( \frac{1}{x} \right) \)

This rule is crucial wherever products of functions are involved in differentiation.

Chain Rule

The chain rule is another essential differentiation tool, especially when dealing with composite functions. A composite function is a function within another function, like \( f(g(x)) \). The chain rule helps find the derivative of such compositions. It states:

• The derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).

In our exercise, consider \( u(x) = \ln(2x) \) as a composite function, specifically \( \ln(g(x)) \) where \( g(x) = 2x \).
Applying the chain rule, find the derivative of \( \ln \) with respect to \( g(x) \) which is \( \frac{1}{g(x)} \), and then multiply by the derivative of \( g(x) \), which is \( 2 \). Therefore, the derivative \( u'(x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \).

• This process is repeated for \( v(x) = \ln(7x) \), resulting in \( v'(x) = \frac{1}{x} \) as well.

The chain rule is powerful in simplifying complex derivatives involving compositions by breaking them into more manageable parts.

Logarithmic Differentiation

Logarithmic differentiation is a technique particularly helpful for functions involving logs and exponents, turning complicated expressions into simpler forms for differentiation. Although not explicitly needed in the primary solution, understanding logarithmic differentiation can enhance comprehension and flexibility with complex problems.
The strategy involves taking the natural logarithm of both sides of an equation, differentiating implicitly, and solving for the derivative. This method is especially beneficial when dealing with products or quotients of functions raised to powers.

• Remember: \( \ln(a^b) = b\ln(a) \)
• Use the identity: \( \ln(a) + \ln(b) = \ln(ab) \)

In the exercise, after applying the product and chain rules, simplifying using \( \ln(2x) + \ln(7x) = \ln(14x^2) \) can be seen as a hint of using logarithmic properties to achieve a tidy form.
Mastering logarithmic differentiation widens the range of functions you can efficiently differentiate and provides a deeper insight into the nature of logarithmic relationships.