Question

find the equation of the line that is tangent to the curve (y = 5xcos x) at the point ((pi,- 5pi)). the equation of this tangent line can be written in the form (y=mx + b) where (m=-5) and (b=)

Explanation:

Step1: Find the derivative of y

The product - rule states that if $y = uv$, where $u = 5x$ and $v=\cos x$, then $y^\prime=u^\prime v + uv^\prime$. We know that $u^\prime = 5$ and $v^\prime=-\sin x$. So $y^\prime=5\cos x-5x\sin x$.

Step2: Evaluate the derivative at $x = \pi$

Substitute $x = \pi$ into $y^\prime$. We have $y^\prime|_{x = \pi}=5\cos\pi-5\pi\sin\pi$. Since $\cos\pi=- 1$ and $\sin\pi = 0$, then $y^\prime|_{x = \pi}=5\times(-1)-5\pi\times0=-5$. This is the slope $m$ of the tangent line.

Step3: Find the value of $b$

We know that the equation of the line is $y = mx + b$, and the line passes through the point $(\pi,-5\pi)$. Substitute $x=\pi$, $y = - 5\pi$ and $m=-5$ into $y=mx + b$. We get $-5\pi=-5\times\pi + b$. Solving for $b$:
\[

$$\begin{align*} -5\pi&=-5\pi + b\\ b&=0 \end{align*}$$

\]

Answer:

$b = 0$