The problem asks for the derivative with respect to $x$ of an integral with an upper limit $x$ and integrand involving $f(x^2 - t^2)$. Let's solve it step-by-step:

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The integral is given as:

$\frac{d}{dx} \int_0^x t f(x^2 - t^2) \, dt$

Step 1: Leibniz Integral Rule

We use the Leibniz rule for differentiation under the integral sign:

$\frac{d}{dx} \int_a^b g(x, t) \, dt = \int_a^b \frac{\partial g(x, t)}{\partial x} \, dt + g(x, b) \cdot \frac{db}{dx} - g(x, a) \cdot \frac{da}{dx}.$

Here:

• $g(x, t) = t f(x^2 - t^2)$,
• $a = 0$ (constant, so $\frac{da}{dx} = 0$),
• $b = x$ (so $\frac{db}{dx} = 1$).

Thus:

$\frac{d}{dx} \int_0^x t f(x^2 - t^2) \, dt = \int_0^x \frac{\partial}{\partial x} \big(t f(x^2 - t^2)\big) \, dt + t f(x^2 - t^2) \big|_{t = x}.$

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Step 2: Compute Partial Derivative

The partial derivative of the integrand $t f(x^2 - t^2)$ with respect to $x$ is:

$\frac{\partial}{\partial x} \big(t f(x^2 - t^2)\big) = t \cdot f'(x^2 - t^2) \cdot \frac{\partial}{\partial x}(x^2 - t^2),$

where:

$\frac{\partial}{\partial x}(x^2 - t^2) = 2x.$

Thus:

$\frac{\partial}{\partial x} \big(t f(x^2 - t^2)\big) = 2x t f'(x^2 - t^2).$

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Step 3: Evaluate the Boundary Term

At $t = x$, the boundary term $t f(x^2 - t^2) \big|_{t = x}$ becomes:

$t f(x^2 - t^2) \big|_{t = x} = x f(x^2 - x^2) = x f(0).$

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Step 4: Combine Terms

The derivative becomes:

$\frac{d}{dx} \int_0^x t f(x^2 - t^2) \, dt = \int_0^x 2x t f'(x^2 - t^2) \, dt + x f(0).$

To match the given options, we consider the case where $f$ is only dependent on $x^2$. This implies:

[ \int_0^x 2x t f'(x^2 - t^2) dt