A function is defined in `(0,oo)` by :
`(x)={(1-x^(2)" for "0ltxle1),(lnx" for "1ltxle2),(ln2-1+0.5x" for "2ltxltoo):}`
Which one of the following is correct in respect of the derivative of the function i.e. f(x) ?

Options

A. `f'(x)=2x" for "0ltxle1`
B. `f'(x)=-2x" for "0ltxle1`
C. `f'(x)=-2x" for "0ltxlt1`
D. `f'(x)=0" for "0ltxltoo`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined in different intervals and find its derivative in each interval. The function is given as:

\[ f(x) = \begin{cases} 1 - x^2 & \text{for } 0 < x \leq 1 \\ \ln x & \text{for } 1 < x \leq 2 \\ \ln 2 - 1 + 0.5x & \text{for } 2 < x < \infty \end{cases} \]

### Step 1: Find the derivative for \( 0 < x \leq 1 \)

For the first interval \( 0 < x \leq 1 \):
\[ f(x) = 1 - x^2 \]
To find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(1 - x^2) = 0 - 2x = -2x \]
Thus, for \( 0 < x \leq 1 \):
\[ f'(x) = -2x \]

### Step 2: Find the derivative for \( 1 < x \leq 2 \)

For the second interval \( 1 < x \leq 2 \):
\[ f(x) = \ln x \]
To find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x} \]
Thus, for \( 1 < x \leq 2 \):
\[ f'(x) = \frac{1}{x} \]

### Step 3: Find the derivative for \( 2 < x < \infty \)

For the third interval \( 2 < x < \infty \):
\[ f(x) = \ln 2 - 1 + 0.5x \]
To find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\ln 2 - 1 + 0.5x) = 0 + 0.5 = 0.5 \]
Thus, for \( 2 < x < \infty \):
\[ f'(x) = 0.5 \]

### Summary of derivatives

We have the following derivatives for each interval:
1. For \( 0 < x \leq 1 \): \( f'(x) = -2x \)
2. For \( 1 < x \leq 2 \): \( f'(x) = \frac{1}{x} \)
3. For \( 2 < x < \infty \): \( f'(x) = 0.5 \)

### Conclusion

Now, we can analyze the options provided in the question regarding the derivative of the function \( f(x) \). The correct answer should reflect the behavior of \( f'(x) \) in each of the defined intervals.