Find the derivatives of the following (1-3) functions :
`f(x)=(ax+b)^(n)`

Answer

To find the derivative of the function \( f(x) = (ax + b)^n \), we will use the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative \( f'(x) \) is given by \( f'(g(x)) \cdot g'(x) \).

### Step-by-Step Solution:

1. **Identify the outer and inner functions**:
- Let \( u = ax + b \) (inner function).
- Then, \( f(x) = u^n \) (outer function).

2. **Differentiate the outer function**:
- The derivative of \( u^n \) with respect to \( u \) is \( n \cdot u^{n-1} \).

3. **Differentiate the inner function**:
- The derivative of \( u = ax + b \) with respect to \( x \) is \( a \).

4. **Apply the chain rule**:
- According to the chain rule, the derivative of \( f(x) \) is:
\[ f'(x) = \frac{d}{dx}(u^n) = n \cdot u^{n-1} \cdot \frac{du}{dx} \]
- Substituting \( u \) and \( \frac{du}{dx} \):
\[ f'(x) = n \cdot (ax + b)^{n-1} \cdot a \]

5. **Final expression for the derivative**:
- Thus, the derivative of \( f(x) = (ax + b)^n \) is:
\[ f'(x) = n \cdot a \cdot (ax + b)^{n-1} \]

### Final Result:
\[ f'(x) = n \cdot a \cdot (ax + b)^{n-1} \]