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.