If `f(x) = {{:(1, x lt 0),(1 + sinx, 0 le x lt pi//2):}` Then at x=0, the derivative `f'(x)` is:

Options

A. 1
B. 0
C. infinite
D. does not exist

Answer

To find the derivative \( f'(x) \) at \( x = 0 \) for the piecewise function

\[ f(x) = \begin{cases} 1 & \text{if } x < 0 \\ 1 + \sin x & \text{if } 0 \leq x < \frac{\pi}{2} \end{cases} \]

we will calculate the left-hand derivative and the right-hand derivative at \( x = 0 \).

### Step 1: Calculate the left-hand derivative at \( x = 0 \)

The left-hand derivative is given by the limit:

\[ f'_-(0) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} \]

Since \( h \) approaches \( 0 \) from the left, \( 0 + h < 0 \), thus \( f(0 + h) = f(h) = 1 \).

Now, we need to find \( f(0) \):

\[ f(0) = 1 + \sin(0) = 1 + 0 = 1 \]

Now substituting into the limit:

\[ f'_-(0) = \lim_{h \to 0^-} \frac{1 - 1}{h} = \lim_{h \to 0^-} \frac{0}{h} = 0 \]

### Step 2: Calculate the right-hand derivative at \( x = 0 \)

The right-hand derivative is given by the limit:

\[ f'_+(0) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} \]

Since \( h \) approaches \( 0 \) from the right, \( 0 + h \geq 0 \), thus \( f(0 + h) = f(h) = 1 + \sin(h) \).

Using \( f(0) = 1 \):

\[ f'_+(0) = \lim_{h \to 0^+} \frac{(1 + \sin(h)) - 1}{h} = \lim_{h \to 0^+} \frac{\sin(h)}{h} \]

We know from the limit property that:

\[ \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \]

Thus,

\[ f'_+(0) = 1 \]

### Step 3: Compare the left-hand and right-hand derivatives

We have:

\[ f'_-(0) = 0 \quad \text{and} \quad f'_+(0) = 1 \]

Since the left-hand derivative does not equal the right-hand derivative:

\[ f'_-(0) \neq f'_+(0) \]

### Conclusion

The derivative \( f'(0) \) does not exist.

### Final Answer

The derivative \( f'(0) \) does not exist.

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