Question

product rule: problem 2
(1 point)
find the equation of the line that is tangent to the curve
y = 3xcos x
at the point ((pi,-3pi)).
the equation of this tangent line can be written in the form (y = mx + b) where
m=
and b=
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Explanation:

Step1: Differentiate using product rule

The product rule states that if $y = uv$, where $u = 3x$ and $v=\cos x$, then $y'=u'v + uv'$. Here, $u' = 3$ and $v'=-\sin x$. So $y'=3\cos x-3x\sin x$.

Step2: Find the slope $m$

Substitute $x = \pi$ into $y'$. $m=y'(\pi)=3\cos\pi-3\pi\sin\pi$. Since $\cos\pi=- 1$ and $\sin\pi = 0$, we have $m=3\times(-1)-3\pi\times0=-3$.

Step3: Find the y - intercept $b$

We know the line is $y = mx + b$, and the line passes through the point $(\pi,-3\pi)$. Substitute $x=\pi$, $y = - 3\pi$ and $m=-3$ into $y=mx + b$. So $-3\pi=-3\pi + b$, which gives $b = 0$.

Answer:

$m=-3$
$b = 0$