If `y=|x-x^(2)|`, then `(dy)/(dx)" at "x=1`.

Options

A. -1
B. 1
C. does not exist
D. none of these

Explanation

To find \(\frac{dy}{dx}\) at \(x = 1\) for the function \(y = |x - x^2|\), we will follow these steps:

### Step 1: Define the function based on the modulus
The expression inside the modulus is \(x - x^2\). We need to determine where this expression is positive or negative to rewrite \(y\) without the modulus.

1. Set \(x - x^2 = 0\):
\[ x(1 - x) = 0 \]
This gives us the roots \(x = 0\) and \(x = 1\).

2. Analyze the sign of \(x - x^2\):
- For \(x < 0\): \(x - x^2 < 0\)
- For \(0 < x < 1\): \(x - x^2 > 0\)
- For \(x > 1\): \(x - x^2 < 0\)

Thus, we can express \(y\) as:
\[ y = \begin{cases} -(x - x^2) & \text{if } x < 0 \text{ or } x > 1 \\ x - x^2 & \text{if } 0 \leq x \leq 1 \end{cases} \]

### Step 2: Differentiate the function
Now we differentiate \(y\) in the two intervals.

1. For \(x < 0\) or \(x > 1\):
\[ y = -(x - x^2) = -x + x^2 \implies \frac{dy}{dx} = -1 + 2x \]

2. For \(0 \leq x \leq 1\):
\[ y = x - x^2 \implies \frac{dy}{dx} = 1 - 2x \]

### Step 3: Evaluate the derivatives at \(x = 1\)
We need to find the left-hand derivative and right-hand derivative at \(x = 1\).

1. **Left-hand derivative** (approaching from the left, \(x \to 1^-\)):
\[ \frac{dy}{dx} = 1 - 2x \text{ at } x = 1 \implies \frac{dy}{dx} = 1 - 2(1) = 1 - 2 = -1 \]

2. **Right-hand derivative** (approaching from the right, \(x \to 1^+\)):
\[ \frac{dy}{dx} = -1 + 2x \text{ at } x = 1 \implies \frac{dy}{dx} = -1 + 2(1) = -1 + 2 = 1 \]

### Step 4: Conclusion
Since the left-hand derivative \(-1\) is not equal to the right-hand derivative \(1\), we conclude that the derivative \(\frac{dy}{dx}\) at \(x = 1\) does not exist.

Thus, the final answer is:
\[ \frac{dy}{dx} \text{ at } x = 1 \text{ does not exist.} \]