Differentiate the function. $$ f(x)=3 x \ln (2 x)-3 x $$ $$ f^{\prime}(x)=\square $$

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Given the function

$$
f(x)=3x\ln(2x)-3x,
$$

we differentiate $$ f(x) $$ term by term.

1. For the first term, use the product rule for $$ 3x\ln(2x) $$. The product rule states that if $$ u(x) $$ and $$ v(x) $$ are functions of $$ x $$, then
   $$
   \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).
   $$
   Let
   $$
   u(x)=3x \quad \text{and} \quad v(x)=\ln(2x).
   $$
   Then,
   $$
   u'(x)=3.
   $$
   To find $$ v'(x) $$, note that
   $$
   \ln(2x) \quad \text{has derivative} \quad \frac{d}{dx}\ln(2x)=\frac{1}{2x}\cdot 2=\frac{1}{x},
   $$
   by the chain rule. Hence, $$ v'(x)=\frac{1}{x} $$.

2. Applying the product rule:
   $$
   \frac{d}{dx}[3x\ln(2x)]=3\ln(2x)+3x\left(\frac{1}{x}\right)=3\ln(2x)+3.
   $$

3. Next, differentiate the second term:
   $$
   \frac{d}{dx}[-3x]=-3.
   $$

4. Combine the derivatives:
   $$
   f'(x)=\left[3\ln(2x)+3\right]-3=3\ln(2x).
   $$

Thus, the derivative is

$$
f'(x)=3\ln(2x).
$$
$$ f'(x) = 3\ln(2x) $$

Differentiate the function. \[ f(x)=3 x \ln (2 x)-3 x \] \( f^{\prime}(x)=\square \)

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