Given: $$ J=(4 x+1) \ln (4 x) $$ (a) Write the name of the rule used to differentiate the given function. (b) Find the derivative $$ \frac{d J}{d x} $$ by showing clearly each step. (c) Calculate the value of $$ \frac{d J}{d x} $$ if $$ x=2 $$. (3 marks)

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(a) The rule used is the **Product Rule**.

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(b) Let 
$$
f(x)=4x+1\quad \text{and}\quad g(x)=\ln(4x).
$$
The product rule states that 
$$
\frac{dJ}{dx} = f'(x)g(x) + f(x)g'(x).
$$

1. Differentiate $$ f(x) $$:
   $$
   f'(x)=4.
   $$

2. Differentiate $$ g(x) $$ using the chain rule. Let $$ u=4x $$ then:
   $$
   g(x)=\ln(u),\quad g'(x)=\frac{d}{dx}\ln(u) = \frac{1}{u}\frac{du}{dx}.
   $$
   Since $$ \frac{du}{dx}=4 $$, we have:
   $$
   g'(x)=\frac{4}{4x}=\frac{1}{x}.
   $$

3. Apply the product rule:
   $$
   \frac{dJ}{dx}=4\ln(4x)+(4x+1)\frac{1}{x}.
   $$
   Simplify the second term:
   $$
   \frac{4x+1}{x} = 4 + \frac{1}{x}.
   $$
   Hence, the derivative is:
   $$
   \frac{dJ}{dx}=4\ln(4x)+4+\frac{1}{x}.
   $$

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(c) Substitute $$ x=2 $$ into the derivative:
$$
\frac{dJ}{dx}\Big|_{x=2}=4\ln(4\cdot2)+4+\frac{1}{2}.
$$
Simplify:
$$
4\cdot 2=8,\quad \text{so} \quad \frac{dJ}{dx}\Big|_{x=2}=4\ln(8)+4+\frac{1}{2}.
$$

Thus, the value of the derivative when $$ x=2 $$ is:
$$
4\ln(8)+4+\frac{1}{2}.
$$
(a) The rule used is the **Product Rule**. (b) The derivative $$ \frac{dJ}{dx} $$ is: $$ \frac{dJ}{dx} = 4\ln(4x) + 4 + \frac{1}{x} $$ (c) When $$ x = 2 $$, the derivative is: $$ \frac{dJ}{dx} = 4\ln(8) + 4 + \frac{1}{2} $$

Given:

(a) Write the name of the rule used to differentiate the given function.
(b) Find the derivative by showing clearly each step.
© Calculate the value of if .
(3 marks)

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Given: (a) Write the name of the rule used to differentiate the given function. (b) Find the derivative by showing clearly each step. © Calculate the value of if . (3 marks)