To find the derivative $f'(x)$ of the function

$f(x) = (7x^5 + 9x^3 - 7) \ln(x),$

we need to apply the product rule and the chain rule for differentiation.

Step-by-step solution:

Let $u(x) = 7x^5 + 9x^3 - 7$ and $v(x) = \ln(x)$.

1. Apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x).$

2. Differentiate $u(x)$:

$u'(x) = \frac{d}{dx}(7x^5 + 9x^3 - 7) = 35x^4 + 27x^2.$

3. Differentiate $v(x)$:

$v'(x) = \frac{d}{dx}(\ln(x)) = \frac{1}{x}.$

4. Substitute into the product rule:

$f'(x) = (35x^4 + 27x^2) \ln(x) + (7x^5 + 9x^3 - 7) \cdot \frac{1}{x}.$

5. Simplify the second term:

$\frac{1}{x}(7x^5 + 9x^3 - 7) = 7x^4 + 9x^2 - \frac{7}{x}.$

Thus, the derivative is:

$f'(x) = (35x^4 + 27x^2) \ln(x) + 7x^4 + 9x^2 - \frac{7}{x}.$

This is the expression for $f'(x)$.

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Would you like further details or have any questions? Here are five related problems you might explore:

1. Differentiate $f(x) = (5x^2 - 4x)\ln(x)$.
2. Find the second derivative of $f(x) = (7x^5 + 9x^3 - 7)\ln(x)$.
3. Compute the derivative of $f(x) = (2x^3 + x)\ln(x^2)$.
4. What is the derivative of $f(x) = x^2 \ln(x^2 + 1)$?
5. Find the derivative of $f(x) = \ln(x^2 + 1)$.

Tip: The product rule is essential when differentiating functions that are products of two functions.