To find the derivative of the function $f(x) = \frac{3}{\sqrt{x^3}}$, let's first rewrite the expression in a form that is easier to differentiate.

Step 1: Rewrite the function

We can rewrite $\frac{3}{\sqrt{x^3}}$ as a power of $x$: $f(x) = 3x^{-\frac{3}{2}}$ since $\sqrt{x^3} = x^{\frac{3}{2}}$ and dividing by $x^{\frac{3}{2}}$ is the same as multiplying by $x^{-\frac{3}{2}}$.

Step 2: Differentiate using the power rule

The power rule for differentiation states that if $f(x) = x^n$, then $f'(x) = n x^{n-1}$.

Now apply the power rule to $f(x) = 3x^{-\frac{3}{2}}$:

$f'(x) = 3 \cdot \left( -\frac{3}{2} \right) x^{-\frac{3}{2} - 1}$ $f'(x) = -\frac{9}{2} x^{-\frac{5}{2}}$

Step 3: Simplify the result

So, the derivative of the function is: $f'(x) = -\frac{9}{2x^{\frac{5}{2}}}$

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What is the second derivative of $f(x) = \frac{3}{\sqrt{x^3}}$?
2. How do you interpret the derivative of this function in terms of rate of change?
3. What happens to $f'(x)$ as $x \to 0$ and $x \to \infty$?
4. How would you compute the derivative of $f(x) = \frac{3}{x^3}$?
5. How can we apply this concept to find the tangent line to $f(x)$ at a specific point?

Tip: When dealing with square roots in functions, it can be helpful to rewrite them as fractional exponents to simplify differentiation.