1. The problem is to find the derivative $f'(x)$ of the function $$f(x) = \int_{e^x}^{2-x} \sin(t^2) \, dt$$ 2. We use the Leibniz rule for differentiation under the integral sign with variable limits: $$f'(x) = \sin((2-x)^2) \cdot \frac{d}{dx}(2-x) - \sin((e^x)^2) \cdot \frac{d}{dx}(e^x)$$ 3. Calculate the derivatives of the limits: $$\frac{d}{dx}(2-x) = -1$$ $$\frac{d}{dx}(e^x) = e^x$$ 4. Substitute these into the formula: $$f'(x) = \sin((2-x)^2) \cdot (-1) - \sin(e^{2x}) \cdot e^x$$ 5. Simplify the expression: $$f'(x) = -\sin((2-x)^2) - e^x \sin(e^{2x})$$ This is the derivative of the given integral function with respect to $x$. Final answer: $$\boxed{f'(x) = -\sin((2-x)^2) - e^x \sin(e^{2x})}$$