The tangent to the curve, y = xeˣ² passing through the point (1, e) also passes through the point :

Q: Wait, why do we use the Product Rule here? Can't we just differentiate $x$ and $e^{x^2}$ separately?

That is a classic trap! The Product Rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ is mandatory because $x$ and $e^{x^2}$ are multiplied together. If you differentiate them separately and multiply the results, you are ignoring the interaction between the two functions. Always respect the structure of the expression!

Q: How do I know if I should use the Chain Rule for $e^{x^2}$?

Whenever you see a function inside another function—like $x^2$ inside the exponential function—the Chain Rule is your best friend. Think of it as peeling an onion: differentiate the outer layer ($e^u$) and then multiply by the derivative of the inner layer ($u = x^2$). It is the key to unlocking complex derivatives.

Q: Why did we substitute $x=1$ only after finding the general derivative?

Great observation! The derivative $\frac{dy}{dx}$ gives you the slope at *any* point $x$. By finding the general expression first, you create a 'slope machine.' Plugging in $x=1$ at the end allows you to find the specific slope at that exact point of tangency. If you plug in numbers too early, you lose the ability to see the function's behavior.

Q: Is there a shortcut to check the options without solving the full line equation?

You could plug each point into the slope formula $\frac{y_2 - y_1}{x_2 - x_1}$ and see if it equals $3e$, but that is actually more work! Finding the equation $y = 3ex - 2e$ is the most robust way to ensure you don't make a calculation error. Trust the process!

Q: What if the point $(1, e)$ wasn't on the curve? Would the tangent still exist?

If the point weren't on the curve, it wouldn't be a 'tangent' to the curve at that point! A tangent must touch the curve at the point of tangency. Always verify that the point satisfies the original equation $y = xe^{x^2}$ before you start differentiating. Here, $1 \cdot e^{1^2} = e$, so we are safe!

Question

The tangent to the curve, $y = xe^{x^2}$ passing through the point $(1, e)$ also passes through the point :

Options:

(A) $(\frac{4}{3}, 2e)$

(B) $(2, 3e)$

(C) $(\frac{5}{3}, 2e)$

(D) $(3, 6e)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a roller coaster track defined by the function $y = xe^{x^2}$. As you move along this path, the steepness of the track changes constantly.

At the specific point $P(1, e)$, you want to know exactly which direction you are headed. This is the essence of the tangent line—it is the straight path you would take if you suddenly lost your grip on the curve and flew off into space.

To find this path, we need two things: a point, which we have as $(1, e)$, and a slope.

### The Calculus Toolkit: Product and Chain Rules

To find the slope, we must calculate the derivative $\frac{dy}{dx}$. Looking at our function $y = xe^{x^2}$, we see a product of two distinct functions: $u = x$ and $v = e^{x^2}$.

The Product Rule, $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$, is our primary weapon here.

But wait, there is a hidden layer! When we differentiate $v = e^{x^2}$, we cannot simply write $e^{x^2}$. We must invoke the Chain Rule.

The derivative of $e^{x^2}$ is $e^{x^2}$ multiplied by the derivative of the exponent, which is $2x$. Thus, $\frac{dv}{dx} = 2x e^{x^2}$.

### Assembling the Derivative

Now, let us bring these pieces together. Substituting our components into the Product Rule, we get:

$$\frac{dy}{dx} = (x)(2x e^{x^2}) + (e^{x^2})(1)$$

By factoring out the common term $e^{x^2}$, the expression simplifies beautifully into:

$$\frac{dy}{dx} = e^{x^2}(2x^2 + 1)$$

This is our 'slope machine.' It tells us the steepness of the curve at any value of $x$. To find the slope at our specific point of interest, $x = 1$, we simply substitute:

$$m = e^{1^2}(2(1)^2 + 1) = e(2 + 1) = 3e$$

### The Final Equation

With the slope $m = 3e$ and the point $(1, e)$ in hand, we use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$. Plugging in our values:

$$y - e = 3e(x - 1)$$

Expanding this, we get $y - e = 3ex - 3e$, which simplifies to the elegant linear equation:

$$y = 3ex - 2e$$

### The Moment of Truth

Now, we test our options. We are looking for a point $(x, y)$ that satisfies this equation. Let us test the point $(\frac{4}{3}, 2e)$.

If we substitute $x = \frac{4}{3}$ into our tangent equation:

$$y = 3e\left(\frac{4}{3}\right) - 2e = 4e - 2e = 2e$$

It matches perfectly! The point $(\frac{4}{3}, 2e)$ lies exactly on the tangent line.

You have successfully navigated the curve, mastered the product rule, and found the path of the tangent. This is the power of calculus—taking a complex, curving reality and finding the simple, linear truth hidden within it.

Animated Solution for Mathematics - Differentiation: The tangent to the curve, $y = x e^{x^{2}}$ passing through the point $(1, e)$ also passes through the point :

Select Answer:

Visualized Solution

Visualizing the Curve $y = x e^{x^{2}}$

The Tool: Product Rule

Differentiating $e^{x^{2}}$

Combining the Terms

Simplifying the Derivative

Calculating Slope at $x = 1$

Final Slope Value

Equation of the Tangent

Simplifying the Equation

Checking the Options

Verifying Option 1

Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus Toolkit

Assembling the Derivative

The Final Equation

The Moment of Truth

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the curve, y=xex2 passing through the point (1,e) also passes through the point :