We are tasked with finding the derivative $\frac{dy}{dx}$ for the function:

$y = \frac{3}{\sqrt{x^3}} - \frac{2}{4\sqrt{x}} + 1$

Let's simplify the terms first:

1. $\frac{3}{\sqrt{x^3}} = 3x^{-3/2}$
2. $\frac{2}{4\sqrt{x}} = \frac{1}{2}x^{-1/2}$

Now the function becomes:

$y = 3x^{-3/2} - \frac{1}{2}x^{-1/2} + 1$

Step 1: Differentiate each term

We will now find the derivative term by term using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.

1. For $3x^{-3/2}$: $\frac{d}{dx}(3x^{-3/2}) = 3 \cdot \left(-\frac{3}{2}\right) x^{-3/2 - 1} = -\frac{9}{2}x^{-5/2}$

2. For $-\frac{1}{2}x^{-1/2}$: $\frac{d}{dx}\left(-\frac{1}{2}x^{-1/2}\right) = -\frac{1}{2} \cdot \left(-\frac{1}{2}\right) x^{-1/2 - 1} = \frac{1}{4}x^{-3/2}$

3. The derivative of the constant $1$ is zero.

Step 2: Combine the results

Thus, the derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -\frac{9}{2}x^{-5/2} + \frac{1}{4}x^{-3/2}$

This is the final expression for $\frac{dy}{dx}$.

Would you like me to explain any step in more detail?

Related Questions:

1. How would you apply the power rule for a function like $x^{n}$ in general?
2. What happens if you differentiate $\frac{dy}{dx}$ again, to get the second derivative?
3. Can you explain why the constant term drops out when differentiating?
4. What would the derivative look like if the function had a higher power of $x$?
5. How do negative exponents affect the derivative?

Tip:

When simplifying expressions involving radicals, convert them to fractional exponents before differentiating to make the process easier.