If f is a periodic function then

Options

A. f' and f" is also periodic
B. f' is periodic but f" is non -periodic
C. f" is periodic but f' is non -periodic
D. None of these

Answer

To solve the problem, we need to show that if \( f \) is a periodic function, then its first and second derivatives \( f' \) and \( f'' \) are also periodic functions.

### Step-by-Step Solution:

1. **Definition of Periodic Function**:
A function \( f(x) \) is said to be periodic if there exists a positive number \( T \) such that \( f(x + T) = f(x) \) for all \( x \). The smallest such \( T \) is called the period of the function.

2. **Assume a Periodic Function**:
Let's assume \( f(x) = \sin x \), which is a well-known periodic function with a period of \( 2\pi \). This means:
\[ f(x + 2\pi) = \sin(x + 2\pi) = \sin x = f(x) \]

3. **Differentiate the Function**:
We differentiate \( f(x) \):
\[ f'(x) = \cos x \]

4. **Check the Period of the First Derivative**:
Now, we need to check if \( f'(x) \) is periodic:
\[ f'(x + 2\pi) = \cos(x + 2\pi) = \cos x = f'(x) \]
Thus, \( f'(x) \) is also periodic with the same period \( 2\pi \).

5. **Differentiate Again**:
Next, we differentiate \( f'(x) \):
\[ f''(x) = -\sin x \]

6. **Check the Period of the Second Derivative**:
We check if \( f''(x) \) is periodic:
\[ f''(x + 2\pi) = -\sin(x + 2\pi) = -\sin x = f''(x) \]
Therefore, \( f''(x) \) is also periodic with the same period \( 2\pi \).

7. **Conclusion**:
From the above steps, we conclude that if \( f(x) \) is a periodic function, then both its first derivative \( f'(x) \) and its second derivative \( f''(x) \) are also periodic functions.