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Use the general power formula to find the derivative of the following functions. Simplify your final answer.

Let's find the derivatives of each function using the power rule and simplify the answers as requested. I'll go through each problem step by step.

### 4. \( y = x^2(x-1)^3(2x-5)^2 \)

To differentiate this, we'll use the product rule along with the chain rule.

Let:
\[ y = x^2 \cdot (x-1)^3 \cdot (2x-5)^2 \]

Using the product rule:
\[ \frac{dy}{dx} = \frac{d}{dx}[x^2] \cdot (x-1)^3(2x-5)^2 + x^2 \cdot \frac{d}{dx}[(x-1)^3(2x-5)^2] \]

1. \( \frac{d}{dx}[x^2] = 2x \)

Now, apply the product rule to \( (x-1)^3(2x-5)^2 \):

Let \( u = (x-1)^3 \) and \( v = (2x-5)^2 \).

Differentiate each:
- \( \frac{du}{dx} = 3(x-1)^2 \)
- \( \frac{dv}{dx} = 4(2x-5) \)

So:
\[
\frac{d}{dx}[(x-1)^3(2x-5)^2] = (3(x-1)^2)(2x-5)^2 + (x-1)^3(4(2x-5))
\]

Now putting everything together:
\[
\frac{dy}{dx} = 2x(x-1)^3(2x-5)^2 + x^2 \left[ 3(x-1)^2(2x-5)^2 + (x-1)^3 4(2x-5) \right]
\]

### 5. \( y = \sqrt{24x - x^2} \)

First, rewrite the square root as a power:
\[
y = (24x - x^2)^{1/2}
\]

Differentiate using the chain rule:
\[
\frac{dy}{dx} = \frac{1}{2}(24x - x^2)^{-1/2} \cdot (24 - 2x)
\]

Simplify:
\[
\frac{dy}{dx} = \frac{(24 - 2x)}{2\sqrt{24x - x^2}}
\]
\[
\frac{dy}{dx} = \frac{12 - x}{\sqrt{24x - x^2}}
\]

### 6. \( y = \frac{(x+1)^3}{x^2} \)

First, rewrite the expression:
\[
y = (x+1)^3 x^{-2}
\]

Now apply the product rule:
\[
\frac{dy}{dx} = 3(x+1)^2 x^{-2} + (x+1)^3 (-2x^{-3})
\]

Simplify:
\[
\frac{dy}{dx} = \frac{3(x+1)^2}{x^2} - \frac{2(x+1)^3}{x^3}
\]

### 7. \( y = \frac{2x^2}{\sqrt{x+1}} \)

First, rewrite \( \sqrt{x+1} \) as \( (x+1)^{1/2} \):
\[
y = 2x^2(x+1)^{-1/2}
\]

Now apply the product rule:
\[
\frac{dy}{dx} = 2(2x)(x+1)^{-1/2} + 2x^2 \cdot \left( -\frac{1}{2}(x+1)^{-3/2} \right)
\]

Simplify:
\[
\frac{dy}{dx} = \frac{4x}{\sqrt{x+1}} - \frac{x^2}{(x+1)^{3/2}}
\]

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These are the derivatives for the given functions. Would you like me to expand on any of the steps, or go further in simplifying some of the answers?

Here are 5 related questions:
1. How do you apply the chain rule in differentiation?
2. What is the product rule, and when is it used?
3. How do you simplify expressions involving roots after differentiation?
4. What are the key steps to apply the power rule in composite functions?
5. How can the quotient rule be used as an alternative approach in these problems?

**Tip**: Always simplify the expression before differentiating, especially when you can apply power rules more easily after rewriting terms.

This page provides a detailed solution for calculating derivatives of functions using the power rule, chain rule, and product rule. The functions include polynomials, square roots, and rational expressions. Ideal for advanced high school or undergraduate math students.

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