The task in the image asks to find the equation of the tangent line to the curve $y = 3x \cos x$ at the point $(\pi, -3\pi)$.

Steps to find the equation of the tangent line:

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

   • The equation of the curve is $y = 3x \cos x$.
   • Using the product rule for differentiation: $\frac{d}{dx}(3x \cos x) = 3 \frac{d}{dx}(x \cos x)$
   • Applying the product rule on $x \cos x$: $\frac{d}{dx}(x \cos x) = \cos x - x \sin x$
   • Therefore, the derivative is: $y' = 3(\cos x - x \sin x)$
   • This derivative gives the slope of the tangent line at any point $x$.

2. Evaluate the slope at $x = \pi$:

   • Plug $x = \pi$ into the derivative: $y'(\pi) = 3(\cos \pi - \pi \sin \pi)$
   • Using $\cos \pi = -1$ and $\sin \pi = 0$: $y'(\pi) = 3(-1 - 0) = -3$
   • So, the slope of the tangent line at $(\pi, -3\pi)$ is $m = -3$.

3. Find the equation of the tangent line:

   • The equation of a line is $y - y_1 = m(x - x_1)$.
   • Here, $m = -3$, $x_1 = \pi$, and $y_1 = -3\pi$.
   • Substituting these values into the point-slope form: $y - (-3\pi) = -3(x - \pi)$ $y + 3\pi = -3x + 3\pi$ $y = -3x + 6\pi$

So, the equation of the tangent line is $y = -3x + 6\pi$, which matches the equation you provided.

Would you like further details or have any questions?

Here are some related questions:

1. What is the product rule and how is it applied?
2. Can we use this approach for finding the tangent to any curve?
3. How do you find the normal line to a curve at a given point?
4. Why do we need the derivative to find the slope of a tangent line?
5. What does the derivative $y'(\pi) = -3$ represent geometrically?

Tip: When differentiating products like $x \cos x$, always remember to apply the product rule: $(uv)' = u'v + uv'$.