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 function \( y = |x| + a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \), we can break it down into two parts: \( y_1 = |x| \) and \( y_2 = a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \).

### Step 1: Differentiate \( y_1 = |x| \)

The absolute value function can be expressed as:
\[ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \]

Thus, the derivative of \( |x| \) is:
\[ \frac{d}{dx} |x| = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0 \end{cases} \]

### Step 2: Differentiate \( y_2 \)

Now, we differentiate \( y_2 \):
\[ y_2 = a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \]

Using the power rule for differentiation, we have:
\[ \frac{dy_2}{dx} = a_0 \cdot n x^{n-1} + a_1 \cdot (n-1) x^{n-2} + a_2 \cdot (n-2) x^{n-3} + \ldots + a_{n-1} \]

The constant term \( a_n \) differentiates to zero.

### Step 3: Combine the derivatives

Now, we combine the derivatives of \( y_1 \) and \( y_2 \):
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]

Thus, we have:
\[ \frac{dy}{dx} = \begin{cases} 1 + (a_0 n x^{n-1} + a_1 (n-1) x^{n-2} + a_2 (n-2) x^{n-3} + \ldots + a_{n-1}) & \text{if } x > 0 \\ -1 + (a_0 n x^{n-1} + a_1 (n-1) x^{n-2} + a_2 (n-2) x^{n-3} + \ldots + a_{n-1}) & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0 \end{cases} \]

### Final Answer

The derivative of the function \( y = |x| + a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \) is:
\[ \frac{dy}{dx} = \begin{cases} 1 + (a_0 n x^{n-1} + a_1 (n-1) x^{n-2} + a_2 (n-2) x^{n-3} + \ldots + a_{n-1}) & \text{if } x > 0 \\ -1 + (a_0 n x^{n-1} + a_1 (n-1) x^{n-2} + a_2 (n-2) x^{n-3} + \ldots + a_{n-1}) & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0 \end{cases} \]