To find the equation of the tangent line to the curve $y = 2x(x^2 - 6x + 10)^8$ at the point $(3, 6)$, we need to follow these steps:

1. Differentiate the function to find the slope of the tangent line.

We'll use the product rule and chain rule for differentiation.

Given: $y = 2x \left( x^2 - 6x + 10 \right)^8$

Let: $u = 2x$ $v = \left( x^2 - 6x + 10 \right)^8$

The product rule states: $\frac{d}{dx} \left( u \cdot v \right) = u' \cdot v + u \cdot v'$

Now, differentiate each part:

• $u' = 2$
• For $v$, apply the chain rule: $v' = 8 \left( x^2 - 6x + 10 \right)^7 \cdot (2x - 6)$

Now, we apply the product rule: $\frac{dy}{dx} = 2 \cdot \left( x^2 - 6x + 10 \right)^8 + 2x \cdot 8 \left( x^2 - 6x + 10 \right)^7 \cdot (2x - 6)$

Simplify this expression: $\frac{dy}{dx} = 2 \left( x^2 - 6x + 10 \right)^8 + 16x(x - 3) \left( x^2 - 6x + 10 \right)^7$

2. Find the slope at $x = 3$.

Now, substitute $x = 3$ into the derivative to get the slope of the tangent line.

First, calculate the value of $\left( x^2 - 6x + 10 \right)$ at $x = 3$: $x^2 - 6x + 10 = 3^2 - 6(3) + 10 = 9 - 18 + 10 = 1$

So: $\frac{dy}{dx} \bigg|_{x=3} = 2(1)^8 + 16(3)(3 - 3)(1)^7 = 2$

The slope of the tangent line at $x = 3$ is 2.

3. Use the point-slope form to write the equation of the tangent line.

The point-slope form of the equation of a line is: $y - y_1 = m(x - x_1)$ where $m$ is the slope, and $(x_1, y_1)$ is the point of tangency.

We have $m = 2$ and the point $(3, 6)$. Substituting these values: $y - 6 = 2(x - 3)$

Simplify: $y - 6 = 2x - 6$ $y = 2x$

Final Answer:

The equation of the tangent line is: $y = 2x$

Would you like any more details or clarification on the process?

Here are some related questions to expand on this topic:

1. How do you find the second derivative of $y = 2x(x^2 - 6x + 10)^8$?
2. What is the equation of the normal line at $x = 3$ for this curve?
3. How does the tangent line behave as $x$ approaches infinity in this function?
4. Can you apply implicit differentiation to curves that are not explicitly defined?
5. What happens to the slope of the tangent line at points where the derivative is zero?

Tip: When using the product rule, carefully separate the functions and apply the derivative to each term to avoid mistakes.