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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


Learn how to find the derivative of an integral with an upper limit x and an integrand involving f(x^2 - t^2) using the Leibniz Integral Rule. The solution includes step-by-step differentiation, boundary term evaluation, and partial derivatives. Suitable for college or advanced students studying calculus.

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