To find the equation of the tangent line to the curve $y = e^{x^3}$ at the point $(0, 1)$, follow these steps:

Step 1: Differentiate the function

To find the slope of the tangent line at a given point, we need to find the derivative of the function $y = e^{x^3}$.

Using the chain rule: $\frac{dy}{dx} = \frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2$ So, the derivative is: $\frac{dy}{dx} = 3x^2 e^{x^3}$

Step 2: Find the slope at the point $(0, 1)$

Substitute $x = 0$ into the derivative to find the slope of the tangent line at the point $(0, 1)$: $\frac{dy}{dx} \bigg|_{x=0} = 3(0)^2 e^{(0)^3} = 3 \cdot 0 \cdot e^0 = 0$ Thus, the slope of the tangent line at $x = 0$ is 0.

Step 3: Equation of the tangent line

The equation of the tangent line is given by the point-slope form: $y - y_1 = m(x - x_1)$ Here, $m = 0$, $(x_1, y_1) = (0, 1)$, so the equation simplifies to: $y - 1 = 0(x - 0)$ $y = 1$

Final Answer:

The equation of the tangent line to the curve $y = e^{x^3}$ at the point $(0, 1)$ is: $y = 1$

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How do you apply the chain rule to more complex functions?
2. What is the general method for finding tangent lines to curves?
3. How would the derivative change if the exponent were different, say $y = e^{x^2}$?
4. What are other examples where the tangent line is horizontal?
5. How does the chain rule relate to implicit differentiation?

Tip: The slope of a tangent line at a point tells you how fast the function is changing at that point. If the slope is zero, the tangent line is horizontal!