We are given the function:

$f(x) = - \frac{\sqrt{x}}{5} + \frac{5 \sqrt{x^3}}{6}$

We need to find the derivative $f'(x)$ and then evaluate it at $x = 4$.

Step 1: Rewrite the function in a more convenient form

First, express the square roots and powers in terms of exponents: $f(x) = -\frac{x^{1/2}}{5} + \frac{5x^{3/2}}{6}$

Step 2: Differentiate the function

Use the power rule for differentiation, $\frac{d}{dx} x^n = n x^{n-1}$, to find $f'(x)$.

For the first term: $\frac{d}{dx} \left( -\frac{x^{1/2}}{5} \right) = -\frac{1}{5} \cdot \frac{1}{2} x^{-1/2} = -\frac{1}{10} x^{-1/2}$

For the second term: $\frac{d}{dx} \left( \frac{5x^{3/2}}{6} \right) = \frac{5}{6} \cdot \frac{3}{2} x^{1/2} = \frac{15}{12} x^{1/2} = \frac{5}{4} x^{1/2}$

Thus, the derivative is: $f'(x) = -\frac{1}{10} x^{-1/2} + \frac{5}{4} x^{1/2}$

Step 3: Evaluate $f'(4)$

Now, substitute $x = 4$ into $f'(x)$:

For the first term: $-\frac{1}{10} \cdot 4^{-1/2} = -\frac{1}{10} \cdot \frac{1}{2} = -\frac{1}{20}$

For the second term: $\frac{5}{4} \cdot 4^{1/2} = \frac{5}{4} \cdot 2 = \frac{10}{4} = \frac{5}{2}$

Thus, $f'(4)$ is: $f'(4) = -\frac{1}{20} + \frac{5}{2}$

Step 4: Combine the terms

To combine the terms, express both terms with a denominator of 20: $f'(4) = -\frac{1}{20} + \frac{50}{20} = \frac{49}{20}$

Final Answer:

$f'(4) = \frac{49}{20}$

Would you like any additional details or clarifications?

Here are five related questions to explore further:

1. How do you differentiate powers of $x$ using the power rule?
2. What is the significance of negative exponents in derivatives?
3. How do square roots convert to fractional exponents?
4. What happens if the power of $x$ is negative when finding the derivative?
5. How would you integrate the given function instead of differentiating it?

Tip: When differentiating functions involving roots and exponents, always convert roots to fractional exponents for easier application of differentiation rules.