Let's walk through the questions based on the image you've uploaded.

Part (a)

The first part asks which logarithmic rule is applied in line (a) of the derivation. The key step is the transformation from:

$\log(f(x)) = \log(a_1(x) \cdot a_2(x) \cdot a_3(x) \cdot a_4(x))$

to:

$\log(f(x)) = \log(a_1(x)) + \log(a_2(x)) + \log(a_3(x)) + \log(a_4(x))$

This is the application of the logarithm of a product rule:

$\log(A \cdot B) = \log(A) + \log(B)$

The correct answer is:
Option 1: $\log(A \cdot B) = \log(A) + \log(B)$

Part (b)

In part (b), we differentiate both sides. The problem states that we use the rule:

$\frac{d}{dx} [\log a(x)] = \frac{1}{a(x)} \cdot a'(x)$

This differentiation involves the chain rule, which allows us to differentiate the composition of functions like $\log a(x)$. The derivative of the logarithmic function involves the reciprocal of the inner function times the derivative of that function.

The correct answer is:
Option 1: Chain rule

Part (c)

For part (c), we need to explain the final step. The derivation simplifies to:

$f'(x) = a_1'(x) \frac{f(x)}{a_1(x)} + a_2'(x) \frac{f(x)}{a_2(x)} + a_3'(x) \frac{f(x)}{a_3(x)} + a_4'(x) \frac{f(x)}{a_4(x)}$

Since $f(x) = a_1(x) \cdot a_2(x) \cdot a_3(x) \cdot a_4(x)$, this allows for some cancellation within each term. Each $f(x)$ term can cancel out with corresponding factors in the denominator.

The final simplification results in terms involving the derivatives of $a_1(x), a_2(x), a_3(x), a_4(x)$.

Would you like any more details or explanations on these steps?

Here are 5 questions to deepen your understanding:

1. How does the chain rule apply in logarithmic differentiation?
2. Could you explain the logarithmic product rule with another example?
3. What are the conditions under which the product rule is used for derivatives?
4. How would the process differ if we used a quotient of functions instead of a product?
5. What are other applications of logarithmic differentiation?

Tip: When differentiating products of multiple functions, consider using logarithmic differentiation if it simplifies the process!