Differentiate the following w.r.t. x:
`|2x^(2)-3|`

Answer

To differentiate the function \( f(x) = |2x^2 - 3| \) with respect to \( x \), we need to consider the points where the expression inside the modulus changes sign. This occurs when \( 2x^2 - 3 = 0 \).

### Step-by-step Solution:

1. **Find the points where the expression inside the modulus is zero:**
\[ 2x^2 - 3 = 0 \implies 2x^2 = 3 \implies x^2 = \frac{3}{2} \implies x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2} \]
Thus, the critical points are \( x = -\frac{\sqrt{6}}{2} \) and \( x = \frac{\sqrt{6}}{2} \).

2. **Determine the intervals based on these points:**
- Interval 1: \( x < -\frac{\sqrt{6}}{2} \)
- Interval 2: \( -\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2} \)
- Interval 3: \( x > \frac{\sqrt{6}}{2} \)

3. **Express \( f(x) \) without the modulus in each interval:**
- For \( x < -\frac{\sqrt{6}}{2} \):
\[ f(x) = 2x^2 - 3 \quad (\text{since } 2x^2 - 3 > 0) \]
- For \( -\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2} \):
\[ f(x) = -(2x^2 - 3) = 3 - 2x^2 \quad (\text{since } 2x^2 - 3 < 0) \]
- For \( x > \frac{\sqrt{6}}{2} \):
\[ f(x) = 2x^2 - 3 \quad (\text{since } 2x^2 - 3 > 0) \]

4. **Differentiate \( f(x) \) in each interval:**
- For \( x < -\frac{\sqrt{6}}{2} \):
\[ f'(x) = \frac{d}{dx}(2x^2 - 3) = 4x \]
- For \( -\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2} \):
\[ f'(x) = \frac{d}{dx}(3 - 2x^2) = -4x \]
- For \( x > \frac{\sqrt{6}}{2} \):
\[ f'(x) = \frac{d}{dx}(2x^2 - 3) = 4x \]

5. **Combine the results:**
\[ f'(x) = \begin{cases} 4x & \text{if } x < -\frac{\sqrt{6}}{2} \\ -4x & \text{if } -\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2} \\ 4x & \text{if } x > \frac{\sqrt{6}}{2} \end{cases} \]