Differentiate `f(x) = (x+2)^(2/3) (1-x)^(1/3)` with respect to `x`

Explanation

To differentiate the function \( f(x) = (x+2)^{2/3} (1-x)^{1/3} \) with respect to \( x \), we will use the product rule. The product rule states that if you have two functions multiplied together, \( u(x) \) and \( v(x) \), then the derivative is given by:

\[ (uv)' = u'v + uv' \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = (x+2)^{2/3} \) and \( v = (1-x)^{1/3} \).

2. **Differentiate \( u \)**:
Using the chain rule, we differentiate \( u \):
\[ u' = \frac{d}{dx} (x+2)^{2/3} = \frac{2}{3}(x+2)^{-1/3} \cdot (1) = \frac{2}{3}(x+2)^{-1/3} \]

3. **Differentiate \( v \)**:
Again using the chain rule, we differentiate \( v \):
\[ v' = \frac{d}{dx} (1-x)^{1/3} = \frac{1}{3}(1-x)^{-2/3} \cdot (-1) = -\frac{1}{3}(1-x)^{-2/3} \]

4. **Apply the product rule**:
Now, we apply the product rule:
\[ f'(x) = u'v + uv' \]
Substituting \( u \), \( u' \), \( v \), and \( v' \):
\[ f'(x) = \left( \frac{2}{3}(x+2)^{-1/3} \right)(1-x)^{1/3} + (x+2)^{2/3}\left( -\frac{1}{3}(1-x)^{-2/3} \right) \]

5. **Simplify the expression**:
We can rewrite the expression:
\[ f'(x) = \frac{2}{3}(x+2)^{-1/3}(1-x)^{1/3} - \frac{1}{3}(x+2)^{2/3}(1-x)^{-2/3} \]

6. **Combine the terms**:
To combine the terms, we can find a common denominator:
\[ f'(x) = \frac{2(1-x) - (x+2)}{3(x+2)^{1/3}(1-x)^{2/3}} \]
Simplifying the numerator:
\[ 2(1-x) - (x+2) = 2 - 2x - x - 2 = -3x \]
Thus, we have:
\[ f'(x) = \frac{-3x}{3(x+2)^{1/3}(1-x)^{2/3}} = \frac{-x}{(x+2)^{1/3}(1-x)^{2/3}} \]

### Final Answer:
\[ f'(x) = \frac{-x}{(x+2)^{1/3}(1-x)^{2/3}} \]