Question:

Step 1: Find the derivative of the curve. We need to find $\frac{dy}{dx}$ for the curve $y = xe^{x^2}$. Using the product rule, where $u = x$ and $v = e^{x^2}$: $$y' = \frac{d}{dx}(x) \cdot e^{x^2} + x \cdot \frac{d}{dx}(e^{x^2})$$ $$y' = e^{x^2} + x \cdot 2xe^{x^2}$$ $$y' = e^{x^2}(1 + 2x^2)$$ Step 2: Calculate the slope of the tangent at the point $(1, e)$. We substitute $x = 1$ into the derivative: $$y'|_{x=1} = e^{1}(1 + 2(1)^2) = e(1 + 2) = 3e$$ Step 3: Write the equation of the tangent line. Using the point-slope form of a line with point $(1, e)$ and slope $m = 3e$: $$y - e = 3e(x - 1)$$ $$y - e = 3ex - 3e$$ $$y = 3ex - 2e$$ Step 4: Check which option satisfies the tangent line equation. We substitute each option into the equation $y = 3ex - 2e$ to see which point lies on the tangent line. For Option 2: $\left(\frac{4}{3}, 2e\right)$ $$y = 3e \cdot \frac{4}{3} - 2e = 4e - 2e = 2e \,\checkmark$$ This matches the $y$-coordinate of the point, confirming it lies on the tangent line. **Final Answer:** The tangent line passes through the point $\left(\frac{4}{3}, 2e\right)$. The correct answer is **Option 2**.