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: Evaluate \( f(0) \)

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

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

Calculating the numerator:

\[ \sin 0 = 0 \quad \text{and} \quad \tan 0 = 0 \]
Thus, the numerator is:

\[ 0 \cdot (0 + 0) = 0 \]

Now, calculating the denominator:

\[ \frac{0 + \pi}{\pi} = 1 \]
So,

\[ \left[\frac{0 + \pi}{\pi}\right] = [1] = 1 \]

Thus, the denominator becomes:

\[ 1 - \frac{1}{2} = \frac{1}{2} \]

Putting it all together:

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

### Step 2: Find \( f'(x) \)

Next, we need to differentiate \( f(x) \) using the quotient rule:

\[ f'(x) = \frac{(u'v - uv')}{v^2} \]

where \( u = x(\sin x + \tan x) \) and \( v = \left[\frac{x + \pi}{\pi}\right] - \frac{1}{2} \).

#### Step 2.1: Differentiate \( u \)

Using the product rule:

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

#### Step 2.2: Differentiate \( v \)

Since \( v \) is a piecewise constant function (the greatest integer function), it will be constant in intervals except at integer points. For small values of \( x \) around 0, \( v = 0.5 \).

Thus, \( v' = 0 \) for \( x \) close to 0.

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

Substituting \( x = 0 \):

\[ u(0) = 0, \quad u'(0) = \sin 0 + \tan 0 + 0 = 0 \]

Since \( v(0) = \frac{1}{2} \) and \( v'(0) = 0 \):

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

### Step 4: Find \( f''(0) \)

To find \( f''(0) \), we differentiate \( f'(x) \) again. Since \( f'(0) = 0 \), we will need to apply the quotient rule again.

However, since both \( u(0) \) and \( u'(0) \) are 0, we can conclude that \( f''(0) \) will depend on the second derivative of \( u \).

#### Step 4.1: Differentiate \( u' \)

We have:

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

Differentiating \( u' \):

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

Evaluating at \( x = 0 \):

\[ u''(0) = 1 + 1 + 1 + 0 = 3 \]

### Final Calculation of \( f''(0) \)

Since \( v(0) = \frac{1}{2} \) and \( v'(0) = 0 \):

Using the quotient rule again:

\[ f''(0) = \frac{u''(0)v(0) - u'(0)v'(0)}{v(0)^2} \]

Substituting values:

\[ f''(0) = \frac{3 \cdot \frac{1}{2} - 0 \cdot 0}{\left(\frac{1}{2}\right)^2} = \frac{\frac{3}{2}}{\frac{1}{4}} = 6 \]

Thus, the final answer is:

\[ \boxed{6} \]