6 Find the derivative of (a) $$ y=(x+2)^{2} \sqrt{x+2} $$ (b) $$ y=\frac{\left(2-3 x^{2}\right)^{2}}{\sqrt{2-3 x^{2}}} $$

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Let's find the derivatives of the given functions step by step.

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### **Problem 6(a)**

**Function:**
$$
y = (x + 2)^2 \sqrt{x + 2}
$$

**Solution:**

1. **Simplify the Function:**
   $$
   y = (x + 2)^2 \cdot (x + 2)^{1/2} = (x + 2)^{2 + \frac{1}{2}} = (x + 2)^{\frac{5}{2}}
   $$

2. **Differentiate Using the Power Rule:**
   $$
   y' = \frac{d}{dx} \left( (x + 2)^{\frac{5}{2}} \right) = \frac{5}{2} (x + 2)^{\frac{5}{2} - 1} = \frac{5}{2} (x + 2)^{\frac{3}{2}}
   $$

**Final Answer for (a):**
$$
y' = \frac{5}{2} (x + 2)^{\frac{3}{2}}
$$

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### **Problem 6(b)**

**Function:**
$$
y = \frac{(2 - 3x^2)^2}{\sqrt{2 - 3x^2}}
$$

**Solution:**

1. **Simplify the Function:**
   $$
   y = \frac{(2 - 3x^2)^2}{(2 - 3x^2)^{\frac{1}{2}}} = (2 - 3x^2)^{2 - \frac{1}{2}} = (2 - 3x^2)^{\frac{3}{2}}
   $$

2. **Differentiate Using the Chain Rule:**
   $$
   y' = \frac{d}{dx} \left( (2 - 3x^2)^{\frac{3}{2}} \right) = \frac{3}{2} (2 - 3x^2)^{\frac{3}{2} - 1} \cdot \frac{d}{dx}(2 - 3x^2)
   $$
   $$
   \frac{d}{dx}(2 - 3x^2) = -6x
   $$
   $$
   y' = \frac{3}{2} (2 - 3x^2)^{\frac{1}{2}} \cdot (-6x) = -9x (2 - 3x^2)^{\frac{1}{2}}
   $$

**Final Answer for (b):**
$$
y' = -9x \sqrt{2 - 3x^2}
$$

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**Summary:**

- **(a)** The derivative is $$ y' = \frac{5}{2} (x + 2)^{\frac{3}{2}} $$.
- **(b)** The derivative is $$ y' = -9x \sqrt{2 - 3x^2} $$.

Answer:
Problem 6(a) Answer

The derivative is $$ y' = \dfrac{5}{2}\, (x + 2)^{\,3/2} $$.

Problem 6(b) Answer

The derivative is $$ y' = -9x\, \sqrt{2 - 3x^{2}} $$.
The derivative of $$ y = (x + 2)^2 \sqrt{x + 2} $$ is $$ y' = \frac{5}{2} (x + 2)^{\frac{3}{2}} $$. The derivative of $$ y = \frac{(2 - 3x^2)^2}{\sqrt{2 - 3x^2}} $$ is $$ y' = -9x \sqrt{2 - 3x^2} $$.

6 Find the derivative of (a) \( y=(x+2)^{2} \sqrt{x+2} \) (b) \( y=\frac{\left(2-3 x^{2}\right)^{2}}{\sqrt{2-3 x^{2}}} \)

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