Find dy/dx for the given function y= 3/(squareroot x^3) - 2/(4sqaureroot x) +1

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.

Find dy/dx for the given function y= 3/(squareroot x^3) - 2/(4sqaureroot x) + 1

Find the derivative of y = 3/√(x^3) - 2/(4√x) + 1 using the power rule and fractional exponents. This step-by-step solution explains how to simplify and differentiate each term in the function. Suitable for students in Grades 11-12 studying calculus.

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