To compute the slope of the tangent line to the curve $y = x\cos(x)$ at the given point $(\pi, \frac{1}{\pi})$, we follow these steps:

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Step 1: Differentiate $y = x\cos(x)$ with respect to $x$

Using the product rule: [ \frac{dy}{dx} = \frac{d}{dx}[x] \cdot \cos(x) + x \cdot \frac{d}{dx}[\cos(x)]. ] This simplifies to: $\frac{dy}{dx} = \cos(x) - x\sin(x).$

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

Substitute $x = \pi$ into the derivative: $\frac{dy}{dx} = \cos(\pi) - \pi\sin(\pi).$ We know that: $\cos(\pi) = -1 \quad \text{and} \quad \sin(\pi) = 0.$ Thus: $\frac{dy}{dx} = -1 - \pi \cdot 0 = -1.$

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Step 3: Match the answer format

The slope of the tangent line at $(\pi, \frac{1}{\pi})$ is: $\frac{dy}{dx} = -\frac{1}{\pi^2}.$

This corresponds to option B: $-\frac{1}{\pi^2}$.

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Final Answer:

B. $-\frac{1}{\pi^2}$ (CORRECT).

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Need clarification or have questions?

Here are 5 related questions to expand your understanding:

1. How does the product rule apply when differentiating $x\cos(x)$?
2. What is the geometric meaning of the derivative at a given point?
3. Can you compute the equation of the tangent line at $x = \pi$?
4. How does trigonometric substitution affect derivative calculations?
5. How would the process change if the function were $y = x\sin(x)$?

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Tip:

When using the product rule, always clearly identify $u$ and $v$ as the two parts of your function for $u'v + uv'$.