Evaluation of Derivative Combination for Twice Differentiable Integral Function

To evaluate the given expression, we first simplify the functional equation for $f(x)$:

$f(x) = \int_0^x \tan (t-x) \, dt - \int_0^x f(t) \tan t \, dt, \quad x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

First, let me evaluate the integral $I_1 = \int_0^x \tan(t-x) \, dt$: $I_1 = [\ln|\sec(t-x)|]_0^x = \ln|\sec(0)| - \ln|\sec(-x)| = 0 - \ln(\sec x) = \ln(\cos x)$

Substituting $I_1$ back into the original equation gives: $f(x) = \ln(\cos x) - \int_0^x f(t) \tan t \, dt$

At $x = 0$, we find the initial condition: $f(0) = \ln(\cos 0) - 0 = 0$

Now, differentiating both sides with respect to $x$ using Leibniz's rule, we get: $f'(x) = \frac{d}{dx}[\ln(\cos x)] - f(x) \tan x$ $f'(x) = -\frac{\sin x}{\cos x} - f(x) \tan x = -\tan x - f(x) \tan x$ $f'(x) = -\tan x \left(1 + f(x)\right)$

This is a separable first-order differential equation: $\frac{f'(x)}{1 + f(x)} = -\tan x$

Integrating both sides with respect to $x$: $\int \frac{f'(x)}{1 + f(x)} \, dx = -\int \tan x \, dx$ $\ln|1 + f(x)| = \ln|\cos x| + C$

Using the initial condition $f(0) = 0$: $\ln|1 + 0| = \ln(\cos 0) + C \implies C = 0$

Since $\cos x > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we have: $1 + f(x) = \cos x \implies f(x) = \cos x - 1$

Now, we calculate the first and second derivatives of $f(x)$: $f'(x) = -\sin x$ $f''(x) = -\cos x$

We need to compute the required expression: $\text{Expression} = f''\left(\frac{\pi}{6}\right) + 12 f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$

Evaluating each term individually:

1. $f''\left(\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
2. $f'\left(-\frac{\pi}{6}\right) = -\sin\left(-\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ $\implies 12 f'\left(-\frac{\pi}{6}\right) = 12 \times \frac{1}{2} = 6$
3. $f\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) - 1 = \frac{\sqrt{3}}{2} - 1$

Summing these values together: $\text{Expression} = \left(-\frac{\sqrt{3}}{2}\right) + 6 + \left(\frac{\sqrt{3}}{2} - 1\right) = 5$