If `int_(0)^(x^(2)) sqrt(1+t^(2))` dt, then f'(x)n equals

Options

A. `sqrt(1+x^(2))`
B. `sqrt(1+x^(4))`
C. `2xsqrt(1+x^(4))`
D. none of these

Answer

To find the derivative of the function defined by the integral

\[ f(x) = \int_{0}^{x^2} \sqrt{1 + t^2} \, dt, \]

we will use the Leibniz rule for differentiation under the integral sign.

### Step-by-Step Solution:

1. **Identify the Function and Limits of Integration:**
The function we have is

\[ f(x) = \int_{0}^{x^2} \sqrt{1 + t^2} \, dt. \]

Here, the lower limit \( u(x) = 0 \) and the upper limit \( v(x) = x^2 \).

2. **Apply the Leibniz Rule:**
According to the Leibniz rule, the derivative of \( f(x) \) is given by:

\[ f'(x) = \sqrt{1 + (v(x))^2} \cdot \frac{d}{dx}[v(x)] - \sqrt{1 + (u(x))^2} \cdot \frac{d}{dx}[u(x)]. \]

Substituting \( u(x) \) and \( v(x) \):

\[ f'(x) = \sqrt{1 + (x^2)^2} \cdot \frac{d}{dx}[x^2] - \sqrt{1 + (0)^2} \cdot \frac{d}{dx}[0]. \]

3. **Calculate the Derivatives:**
- The derivative of \( v(x) = x^2 \) is \( \frac{d}{dx}[x^2] = 2x \).
- The derivative of \( u(x) = 0 \) is \( \frac{d}{dx}[0] = 0 \).

4. **Substitute and Simplify:**
Now substituting these derivatives back into the equation:

\[ f'(x) = \sqrt{1 + x^4} \cdot (2x) - \sqrt{1 + 0} \cdot 0. \]

This simplifies to:

\[ f'(x) = 2x \sqrt{1 + x^4}. \]

5. **Final Result:**
Thus, the derivative \( f'(x) \) is:

\[ f'(x) = 2x \sqrt{1 + x^4}. \]