To find the equation of the tangent line to the curve $y = x^6 \cos(x)$ at the point $(\pi, -\pi^6)$, we need to follow these steps:

Step 1: Find the derivative of the function

The equation of the tangent line requires the slope of the curve at the given point, which is obtained from the derivative $\frac{dy}{dx}$.

The function is $y = x^6 \cos(x)$. To differentiate this, we will use the product rule, which states: $\frac{d}{dx} [f(x) g(x)] = f'(x) g(x) + f(x) g'(x)$ Let $f(x) = x^6$ and $g(x) = \cos(x)$. Now, differentiate each part:

• $f'(x) = 6x^5$
• $g'(x) = -\sin(x)$

Using the product rule: $\frac{dy}{dx} = 6x^5 \cos(x) + x^6 (-\sin(x))$ $\frac{dy}{dx} = 6x^5 \cos(x) - x^6 \sin(x)$

Step 2: Evaluate the derivative at $x = \pi$

To find the slope of the tangent line at the point $x = \pi$, substitute $x = \pi$ into the derivative:

$\frac{dy}{dx}\bigg|_{x = \pi} = 6\pi^5 \cos(\pi) - \pi^6 \sin(\pi)$

We know that:

• $\cos(\pi) = -1$
• $\sin(\pi) = 0$

Thus: $\frac{dy}{dx}\bigg|_{x = \pi} = 6\pi^5 (-1) - \pi^6 (0) = -6\pi^5$

So, the slope of the tangent line is $m = -6\pi^5$.

Step 3: Use the point-slope form to find the equation of the tangent line

The point-slope form 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 on the line.

Using the point $(\pi, -\pi^6)$ and the slope $m = -6\pi^5$, we substitute into the point-slope form: $y - (-\pi^6) = -6\pi^5 (x - \pi)$ Simplifying: $y + \pi^6 = -6\pi^5 (x - \pi)$

This is the equation of the tangent line to the curve $y = x^6 \cos(x)$ at the point $(\pi, -\pi^6)$.

Would you like more details on any specific part of the process or further clarification?

Here are 5 questions for deeper exploration:

1. How do you apply the product rule in differentiation?
2. What is the significance of the point-slope form of a line?
3. Why does $\cos(\pi)$ equal $-1$ and $\sin(\pi)$ equal $0$?
4. How do you interpret the derivative in terms of the curve's slope at a point?
5. What are some real-world applications of finding tangent lines?

Tip: Remember that the product rule is useful whenever you are differentiating a product of two functions.