Find an equation of the tangent line to the curve $y = x\cos(x)$ at the point $(\pi, -\pi)$ I concluded that the equation is
$$(y + \pi) = (\cos(x) + x-\sin(x)) (x - \pi)$$
1) Is this correct so far?  Wolfram doesn't seem to process this correctly.
2) How would I expand this to get it in $y$-intercept form? I know I can plug $\pi$ into my derivative, but i'm not sure what to make of plugging $x$ into it. I feel like it wouldn't be correct, or it would be correct but it would be so verbose the data would be unusable. 

Recall that the product rule is
$$\dfrac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$$
Let $f(x) = x$ and $g(x) = \cos(x)$.  Then, $f'(x) = 1$ and $g'(x) = -\sin(x)$.  So the derivative of $y = x\cos(x)$ is
$$y' = \cos(x) - x\sin(x)$$
Then, at $x = \pi$
$$y'(\pi) = \cos(\pi) - \pi \cdot \sin(\pi) = -1 - \pi \cdot 0 = -1$$
Thus, the equation of the tangent line is
$$y + \pi = -(x - \pi)$$
