The problem asks to find the equation of the tangent line to the curve $y = x^4 + 6e^x$ at the point $(0, 6)$.

To find the equation of the tangent line, we follow these steps:

Step 1: Calculate the derivative $\frac{dy}{dx}$.

The derivative of the function $y = x^4 + 6e^x$ gives us the slope of the tangent line. Using standard rules of differentiation:

$\frac{dy}{dx} = \frac{d}{dx}(x^4) + \frac{d}{dx}(6e^x)$ $\frac{dy}{dx} = 4x^3 + 6e^x$

Step 2: Evaluate the derivative at $x = 0$.

To find the slope of the tangent line at the point $(0, 6)$, we substitute $x = 0$ into the derivative:

$\frac{dy}{dx}\bigg|_{x=0} = 4(0)^3 + 6e^0 = 6$

So, the slope of the tangent line at $x = 0$ is 6.

Step 3: Use the point-slope form of a line.

The point-slope form of the equation of a line is given by:

$y - y_1 = m(x - x_1)$

Where $m$ is the slope, and $(x_1, y_1)$ is the point of tangency. In this case, the slope $m = 6$ and the point is $(0, 6)$.

Substitute these values into the equation:

$y - 6 = 6(x - 0)$

Simplifying:

$y = 6x + 6$

Thus, the equation of the tangent line is: $y = 6x + 6$

Would you like more details on any step or have further questions?

Here are 5 related questions to explore:

1. How do you find the second derivative of $y = x^4 + 6e^x$?
2. What is the general method to find a tangent line for any given function at a point?
3. How would you find the normal line to this curve at the same point?
4. What is the geometric interpretation of the derivative at a specific point?
5. Can we approximate $y = x^4 + 6e^x$ around $x = 0$ using Taylor series?

Tip: Remember that the tangent line provides the best linear approximation of a curve at a specific point.