To find the derivative $f'(x)$ of the function $f(x) = 7x^3 \ln x$, we will apply the product rule because the function is a product of two functions: $7x^3$ and $\ln x$.

Steps:

1. Product Rule: If $f(x) = u(x) v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

   • Here, $u(x) = 7x^3$ and $v(x) = \ln x$.

2. Differentiate each function:

   • $u'(x) = \frac{d}{dx} [7x^3] = 21x^2$.
   • $v'(x) = \frac{d}{dx} [\ln x] = \frac{1}{x}$.

3. Apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ $f'(x) = (21x^2)(\ln x) + (7x^3)\left(\frac{1}{x}\right)$

4. Simplify: $f'(x) = 21x^2 \ln x + 7x^2$

Thus, the derivative is: $f'(x) = 21x^2 \ln x + 7x^2$

Would you like further details or have any questions?

Here are five related questions to expand on this concept:

1. How would the derivative change if the function were $f(x) = 7x^3 \ln(x^2)$?
2. What happens if you replace $\ln x$ with $\log x$ in the original function?
3. Can you apply the product rule to a function like $f(x) = e^x \ln x$?
4. How would you find the second derivative, $f''(x)$, of the same function?
5. What is the physical interpretation of this derivative in a real-world context?

Tip: Always simplify each part of a product or quotient before applying differentiation rules to avoid errors.