Let `f(x)={{:((1+sinx)",","when "0lexlt(pi)/(2)),(1","," when "xlt0):}` . Find Rf'(0)

Answer

To find the right-hand derivative \( f'(0) \) for the function defined as:

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

we will follow these steps:

### Step 1: Identify the right-hand derivative formula
The right-hand derivative at \( x = 0 \) is given by the limit:

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

### Step 2: Calculate \( f(0) \)
Since \( 0 \) is in the interval \( [0, \frac{\pi}{2}) \), we use the first case of the function:

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

### Step 3: Substitute into the derivative formula
Now we substitute \( f(0) \) into the derivative formula:

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

### Step 4: Determine \( f(h) \) for \( h > 0 \)
For \( h > 0 \), since \( h \) is still in the interval \( [0, \frac{\pi}{2}) \), we have:

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

### Step 5: Substitute \( f(h) \) into the limit
Now we can substitute \( f(h) \) into our limit:

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

### Step 6: Evaluate the limit
We know from calculus that:

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

Thus, we have:

\[ f'(0) = 1 \]

### Final Answer
The right-hand derivative \( f'(0) \) is:

\[ \boxed{1} \]