Let `f(x)=(x(sinx+tanx))/([(x+pi)/(pi)]-1//2)` (where (.] denotes the greatest integer function) then find `f"(0)`.

Answer

To find \( f''(0) \) for the function

\[ f(x) = \frac{x(\sin x + \tan x)}{\left(\frac{x + \pi}{\pi}\right) - \frac{1}{2}} \]

we will follow these steps:

### Step 1: Simplify the function at \( x = 0 \)

First, we need to evaluate \( f(0) \):

\[ f(0) = \frac{0(\sin 0 + \tan 0)}{\left(\frac{0 + \pi}{\pi}\right) - \frac{1}{2}} \]

Calculating the numerator:

\[ 0(\sin 0 + \tan 0) = 0 \]

Now, calculating the denominator:

\[ \frac{0 + \pi}{\pi} - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2} \]

Thus,

\[ f(0) = \frac{0}{\frac{1}{2}} = 0 \]

### Step 2: Find the first derivative \( f'(x) \)

To find \( f'(x) \), we will use the quotient rule:

\[ f'(x) = \frac{g(x)h'(x) - h(x)g'(x)}{(h(x))^2} \]

where \( g(x) = x(\sin x + \tan x) \) and \( h(x) = \left(\frac{x + \pi}{\pi}\right) - \frac{1}{2} \).

Calculating \( h(x) \):

\[ h(x) = \frac{x + \pi}{\pi} - \frac{1}{2} = \frac{x + \pi - \frac{\pi}{2}}{\pi} = \frac{x + \frac{\pi}{2}}{\pi} \]

Now, calculate \( h'(x) \):

\[ h'(x) = \frac{1}{\pi} \]

Next, we need \( g'(x) \):

Using the product rule:

\[ g'(x) = (\sin x + \tan x) + x(\cos x + \sec^2 x) \]

### Step 3: Evaluate \( f'(0) \)

Now substituting \( x = 0 \):

\[ g(0) = 0(\sin 0 + \tan 0) = 0 \]

\[ h(0) = \frac{0 + \frac{\pi}{2}}{\pi} = \frac{1}{2} \]

Thus,

\[ f'(0) = \frac{0 \cdot h'(0) - h(0) \cdot g'(0)}{(h(0))^2} \]

Calculating \( g'(0) \):

\[ g'(0) = (\sin 0 + \tan 0) + 0(\cos 0 + \sec^2 0) = 0 + 0 = 0 \]

So,

\[ f'(0) = \frac{0 - \frac{1}{2} \cdot 0}{\left(\frac{1}{2}\right)^2} = 0 \]

### Step 4: Find the second derivative \( f''(x) \)

To find \( f''(x) \), we differentiate \( f'(x) \) again using the quotient rule. However, since \( f'(0) = 0 \), we can directly evaluate the limit to find \( f''(0) \).

### Step 5: Evaluate \( f''(0) \)

Since both \( f(0) \) and \( f'(0) \) are zero, we can use L'Hôpital's rule to find \( f''(0) \):

\[ f''(0) = \lim_{x \to 0} \frac{f'(x)}{x} \]

As we have already established that \( f'(0) = 0 \), we differentiate \( f'(x) \) again and evaluate at \( x = 0 \).

### Final Result

After evaluating the derivatives, we find:

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