Differentiate `|x|+a_(0)x^(n)+a_(1)x^(n-1)+a_(2)x^(n-1)+...+a_(n-1)x+a_(n)`

Answer

To differentiate the expression \( |x| + a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \), we need to consider the absolute value function \( |x| \) separately for the cases when \( x < 0 \) and \( x \geq 0 \).

### Step-by-step Solution:

1. **Identify the cases for \( |x| \)**:
- For \( x < 0 \), \( |x| = -x \).
- For \( x \geq 0 \), \( |x| = x \).

2. **Write the function \( y \)**:
- For \( x < 0 \):
\[ y = -x + a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_n \]
- For \( x \geq 0 \):
\[ y = x + a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_n \]

3. **Differentiate for \( x < 0 \)**:
- Differentiate \( y \):
\[ \frac{dy}{dx} = -1 + n a_0 x^{n-1} + (n-1) a_1 x^{n-2} + (n-2) a_2 x^{n-3} + \ldots + 0 \]
- This simplifies to:
\[ \frac{dy}{dx} = -1 + n a_0 x^{n-1} + (n-1) a_1 x^{n-2} + \ldots + a_1 \]

4. **Differentiate for \( x \geq 0 \)**:
- Differentiate \( y \):
\[ \frac{dy}{dx} = 1 + n a_0 x^{n-1} + (n-1) a_1 x^{n-2} + (n-2) a_2 x^{n-3} + \ldots + 0 \]
- This simplifies to:
\[ \frac{dy}{dx} = 1 + n a_0 x^{n-1} + (n-1) a_1 x^{n-2} + \ldots + a_1 \]

5. **Combine the results**:
- The derivative can be expressed as:
\[ \frac{dy}{dx} = \begin{cases} -1 + n a_0 x^{n-1} + (n-1) a_1 x^{n-2} + \ldots & \text{if } x < 0 \\ 1 + n a_0 x^{n-1} + (n-1) a_1 x^{n-2} + \ldots & \text{if } x \geq 0 \end{cases} \]