If `y = underset(u(x))overset(v(x))intf(t) dt`, let us define `(dy)/(dx)` in a different manner as `(dy)/(dx) = v'(x) f^(2)(v(x)) - u'(x) f^(2)(u(x))` alnd the equation of the tangent at `(a,b)` as `y -b = (dy/dx)_((a,b)) (x-a)`
If `F(x) = underset(1)overset(x)inte^(t^(2)//2)(1-t^(2))dt`, then `d/(dx) F(x)` at `x = 1` is

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

To solve the problem, we need to find the derivative of the function \( F(x) = \int_{1}^{x} e^{\frac{t^2}{2}} (1 - t^2) dt \) at \( x = 1 \).

### Step-by-Step Solution:

1. **Identify the Function**:
We have \( F(x) = \int_{1}^{x} e^{\frac{t^2}{2}} (1 - t^2) dt \).

2. **Apply the Leibniz Rule**:
According to the Leibniz rule for differentiation under the integral sign, we can differentiate \( F(x) \) as follows:
\[ \frac{d}{dx} F(x) = e^{\frac{x^2}{2}} (1 - x^2) \cdot \frac{d}{dx}(x) - e^{\frac{1^2}{2}} (1 - 1^2) \cdot \frac{d}{dx}(1) \]
Here, the first term corresponds to the upper limit \( x \) and the second term corresponds to the lower limit \( 1 \).

3. **Differentiate the Upper Limit**:
The derivative of the upper limit \( x \) is \( 1 \). Therefore:
\[ \frac{d}{dx} F(x) = e^{\frac{x^2}{2}} (1 - x^2) \]

4. **Differentiate the Lower Limit**:
The derivative of the lower limit \( 1 \) is \( 0 \), so the second term becomes \( 0 \).

5. **Evaluate at \( x = 1 \)**:
Now, we need to evaluate \( \frac{d}{dx} F(x) \) at \( x = 1 \):
\[ \frac{d}{dx} F(1) = e^{\frac{1^2}{2}} (1 - 1^2) = e^{\frac{1}{2}} (1 - 1) = e^{\frac{1}{2}} \cdot 0 = 0 \]

6. **Final Result**:
Thus, the value of \( \frac{d}{dx} F(x) \) at \( x = 1 \) is:
\[ \frac{d}{dx} F(1) = 0 \]

### Summary:
The derivative \( \frac{d}{dx} F(x) \) at \( x = 1 \) is \( 0 \).