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The function given is:

\[
f(x) = 5^x \left( -3x^4 + \frac{4}{x} \right)
\]

To find the derivative \( f'(x) \), we need to use the **product rule**, which states:

\[
\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'
\]

Let:
- \( u(x) = 5^x \), and
- \( v(x) = -3x^4 + \frac{4}{x} \).

### Step 1: Derivative of \( u(x) = 5^x \)
Using the chain rule and the fact that \( \frac{d}{dx} a^x = a^x \ln a \):

\[
u'(x) = 5^x \ln 5
\]

### Step 2: Derivative of \( v(x) = -3x^4 + \frac{4}{x} \)
We differentiate term by term:

- \( \frac{d}{dx}(-3x^4) = -12x^3 \),
- \( \frac{d}{dx} \left( \frac{4}{x} \right) = -\frac{4}{x^2} \).

Thus, 
\[
v'(x) = -12x^3 - \frac{4}{x^2}
\]

### Step 3: Apply the product rule
Now, apply the product rule:

\[
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

Substituting the expressions for \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \):

\[
f'(x) = 5^x \ln 5 \left( -3x^4 + \frac{4}{x} \right) + 5^x \left( -12x^3 - \frac{4}{x^2} \right)
\]

### Final result:

\[
f'(x) = 5^x \left[ \ln 5 \left( -3x^4 + \frac{4}{x} \right) + \left( -12x^3 - \frac{4}{x^2} \right) \right]
\]

Would you like further clarification or details on any part of this process?

### Related Questions:
1. What is the chain rule and how does it apply to functions of the form \( a^x \)?
2. How do we differentiate functions that are products of two terms?
3. Why is the product rule necessary when differentiating a product of functions?
4. What is the logarithmic derivative, and how is it applied to exponential functions?
5. How can we simplify the expression after applying the product rule?

**Tip:** When differentiating products, always identify and differentiate each part (product rule) before multiplying the results.

Find the derivative of f(x) = 5^x (-3x^4 + 4/x).

This detailed solution explains how to differentiate the function f(x) = 5^x (-3x^4 + 4/x) using the product rule and chain rule, making it ideal for students studying calculus at Grades 11-12 or college level.

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