To find the equation of the line tangent to the curve $y = 3x \cos x$ at the point $(\pi, -3\pi)$, we need to follow these steps:

Step 1: Differentiate $y = 3x \cos x$

We start by finding the derivative of $y$ with respect to $x$ to get the slope of the tangent line. Using the product rule, where $u = 3x$ and $v = \cos x$, we have:

$\frac{dy}{dx} = u'v + uv'$

Calculating each term:

• $u = 3x \Rightarrow u' = 3$
• $v = \cos x \Rightarrow v' = -\sin x$

So, $\frac{dy}{dx} = (3)(\cos x) + (3x)(-\sin x) = 3 \cos x - 3x \sin x$

Step 2: Substitute $x = \pi$ into $\frac{dy}{dx}$

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

$\frac{dy}{dx} \bigg|_{x=\pi} = 3 \cos(\pi) - 3\pi \sin(\pi)$

Since $\cos(\pi) = -1$ and $\sin(\pi) = 0$, this simplifies to:

$\frac{dy}{dx} \bigg|_{x=\pi} = 3(-1) - 3\pi(0) = -3$

So, the slope of the tangent line at $x = \pi$ is $-3$.

Step 3: Write the equation of the tangent line

The point of tangency is $(\pi, -3\pi)$, and the slope $m = -3$. Using the point-slope form of a line $y - y_1 = m(x - x_1)$:

$y - (-3\pi) = -3(x - \pi)$

Simplifying, we get:

$y + 3\pi = -3x + 3\pi$

Subtracting $3\pi$ from both sides:

$y = -3x$

Final Answer:

The equation of the tangent line at $(\pi, -3\pi)$ is:

$y = -3x$

Would you like more details or have any questions?

Here are five related questions:

1. How do you find the equation of a tangent line for different types of functions?
2. What is the importance of the product rule in calculus?
3. How does the point-slope form of a line help in finding tangent lines?
4. Can tangent lines be vertical, and if so, under what conditions?
5. How would you find the tangent line if given a parametric curve?

Tip: Knowing the fundamental differentiation rules (like product, quotient, and chain rules) is essential for tackling calculus problems involving slopes and tangent lines.