Key Concepts and Formulas

• Derivative as Slope: The derivative of a function $y = f(x)$, denoted as $\frac{dy}{dx}$ or $f'(x)$, gives the slope of the tangent line to the curve at any point $x$.
• Equation of a Tangent Line: The equation of the tangent line to the curve $y = f(x)$ at the point $(x_1, y_1)$ is given by the point-slope form: $y - y_1 = m(x - x_1)$, where $m$ is the slope of the tangent at that point, i.e., $m = f'(x_1)$.
• Product Rule: If $y = u(x)v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$.
• Chain Rule: If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

Step-by-Step Solution

Step 1: Verify the Point Lies on the Curve

The given curve is $y = xe^{x^2}$, and the point is $(1, e)$. We need to ensure that the point $(1, e)$ indeed lies on the curve. Substituting $x = 1$ into the equation of the curve: $y = (1)e^{(1)^2} = e^1 = e$ Since $y = e$ when $x = 1$, the point $(1, e)$ lies on the curve.

Step 2: Differentiate the Curve

We need to find the derivative of $y = xe^{x^2}$ with respect to $x$ to determine the slope of the tangent at any point on the curve. We will use the product rule, where $u(x) = x$ and $v(x) = e^{x^2}$. $u'(x) = \frac{d}{dx}(x) = 1$ To find $v'(x)$, we use the chain rule. Let $z = x^2$. Then $v(x) = e^z$, and $\frac{dv}{dx} = \frac{dv}{dz} \cdot \frac{dz}{dx}$. $\frac{dv}{dz} = \frac{d}{dz}(e^z) = e^z = e^{x^2}$ $\frac{dz}{dx} = \frac{d}{dx}(x^2) = 2x$ Therefore, $v'(x) = e^{x^2} \cdot 2x$. Now, applying the product rule: $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = (1)e^{x^2} + x(e^{x^2} \cdot 2x) = e^{x^2} + 2x^2e^{x^2} = e^{x^2}(1 + 2x^2)$. Thus, $\frac{dy}{dx} = e^{x^2}(1 + 2x^2)$.

Step 3: Calculate the Slope at the Point (1, e)

We substitute $x = 1$ into the derivative $\frac{dy}{dx}$ to find the slope of the tangent at the point $(1, e)$. $m = \left. \frac{dy}{dx} \right|_{x=1} = e^{(1)^2}(1 + 2(1)^2) = e(1 + 2) = 3e$. So, the slope of the tangent line at $(1, e)$ is $3e$.

Step 4: Formulate the Equation of the Tangent Line

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (1, e)$ and $m = 3e$: $y - e = 3e(x - 1)$ $y - e = 3ex - 3e$ $y = 3ex - 3e + e$ $y = 3ex - 2e$ This is the equation of the tangent line.

Step 5: Verify Which Option Lies on the Tangent Line

We need to check which of the given options satisfies the equation $y = 3ex - 2e$.

Let's check option (A): $\left( \frac{4}{3}, 2e \right)$ Substitute $x = \frac{4}{3}$ and $y = 2e$ into the tangent equation: $2e = 3e \left( \frac{4}{3} \right) - 2e$ $2e = 4e - 2e$ $2e = 2e$ Since the equation holds true, the point $\left( \frac{4}{3}, 2e \right)$ lies on the tangent line.

Common Mistakes & Tips

• Chain Rule Application: Remember to apply the chain rule correctly when differentiating composite functions like $e^{x^2}$. Neglecting the inner derivative ($2x$ in this case) is a common error.
• Product Rule Application: When applying the product rule, make sure to correctly identify the two functions being multiplied and their derivatives.
• Algebraic Errors: Be careful with algebraic manipulations, especially when simplifying the equation of the tangent line.

Summary

To find the tangent to the curve $y = xe^{x^2}$ at the point $(1, e)$ that also passes through another point, we first verified that the point lies on the curve, then found the derivative of the curve to determine the slope of the tangent. Then, we used the point-slope form to derive the tangent line equation, and finally, we checked each of the given options to see which one satisfies the tangent equation. The point $\left( \frac{4}{3}, 2e \right)$ satisfies the equation.

The final answer is $\boxed{\left( {{4 \over 3},2e} \right)}$, which corresponds to option (A).