Find the derivative of $$ f(x)=x^{3} 7^{3 x} $$ $$ f^{\prime}(x)=\square $$

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We are given 
$$
f(x)=x^{3} \cdot 7^{3x}.
$$
To find the derivative $$ f'(x) $$, we use the product rule:
$$
f'(x)=\frac{d}{dx}(x^{3}) \cdot 7^{3x}+ x^{3} \cdot \frac{d}{dx}(7^{3x}).
$$

**Step 1:** Differentiate $$ x^{3} $$.

$$
\frac{d}{dx}(x^{3}) = 3x^{2}.
$$

**Step 2:** Differentiate $$ 7^{3x} $$.

Recall the rule for differentiating an exponential function $$ a^{u(x)} $$:
$$
\frac{d}{dx}\left(a^{u(x)}\right)=a^{u(x)}\ln(a)\cdot u'(x).
$$
Here, $$ a=7 $$ and $$ u(x)=3x $$ with $$ u'(x)=3 $$. Hence,
$$
\frac{d}{dx}(7^{3x}) = 7^{3x}\ln(7)\cdot 3 = 3\ln(7) \cdot 7^{3x}.
$$

**Step 3:** Combine the results using the product rule.

$$
f'(x) = 3x^{2}\cdot 7^{3x} + x^{3}\cdot \left(3\ln(7) \cdot 7^{3x}\right).
$$

Factor out common factors $$ 3 \cdot 7^{3x} $$:

$$
f'(x)=3\cdot 7^{3x}\left(x^{2}+ x^{3}\ln(7)\right).
$$

Thus, the derivative is

$$
f'(x)=3 \cdot 7^{3x}\left(x^{2}+ x^{3}\ln(7)\right).
$$
The derivative of $$ f(x) = x^{3} \cdot 7^{3x} $$ is $$ f'(x) = 3 \cdot 7^{3x} \left( x^{2} + x^{3} \ln(7) \right) $$.

Find the derivative of \( f(x)=x^{3} 7^{3 x} \) \( f^{\prime}(x)=\square \)

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