Let `f(x)` be a differentiable function satisfying `f(x)=int_(0)^(x)e^((2tx-t^(2)))cos(x-t)dt`, then find the value of `f''(0)`.

Explanation

To find the value of \( f''(0) \) for the function defined as

\[ f(x) = \int_{0}^{x} e^{(2tx - t^2)} \cos(x - t) \, dt, \]

we will follow these steps:

### Step 1: Differentiate \( f(x) \)

Using the Leibniz rule for differentiation under the integral sign, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \int_{0}^{x} e^{(2tx - t^2)} \cos(x - t) \, dt \right) = e^{(2xx - x^2)} \cos(0) + \int_{0}^{x} \frac{\partial}{\partial x} \left( e^{(2tx - t^2)} \cos(x - t) \right) dt. \]

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

The first term simplifies to:

\[ e^{(2x^2 - x^2)} \cdot 1 = e^{x^2}. \]

Now we need to compute the integral term:

\[ \int_{0}^{x} \left( 2te^{(2tx - t^2)} \cos(x - t) - e^{(2tx - t^2)} \sin(x - t) \right) dt. \]

Thus, we have:

\[ f'(x) = e^{x^2} + \int_{0}^{x} \left( 2te^{(2tx - t^2)} \cos(x - t) - e^{(2tx - t^2)} \sin(x - t) \right) dt. \]

### Step 3: Differentiate \( f'(x) \) to find \( f''(x) \)

Next, we differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx} \left( e^{x^2} \right) + \frac{d}{dx} \left( \int_{0}^{x} \left( 2te^{(2tx - t^2)} \cos(x - t) - e^{(2tx - t^2)} \sin(x - t) \right) dt \right). \]

The derivative of \( e^{x^2} \) is:

\[ 2xe^{x^2}. \]

Using the Leibniz rule again for the integral:

\[ f''(x) = 2xe^{x^2} + \left( 2xe^{(2xx - x^2)} \cos(0) - e^{(2xx - x^2)} \sin(0) \right) + \int_{0}^{x} \frac{\partial}{\partial x} \left( 2te^{(2tx - t^2)} \cos(x - t) - e^{(2tx - t^2)} \sin(x - t) \right) dt. \]

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

Now, we substitute \( x = 0 \):

- The term \( 2xe^{x^2} \) becomes \( 0 \).
- The first term from the integral also becomes \( 0 \) since both upper and lower limits are the same.

Thus, we have:

\[ f''(0) = 2 \cdot 0 \cdot e^{0} + 0 + 0 = 0. \]

### Final Answer

Therefore, the value of \( f''(0) \) is

\[ \boxed{0}. \]