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=) and (b=)

Explanation:

Step1: Find the derivative of $y = 5x\cos x$

Use the product - rule $(uv)^\prime=u^\prime v + uv^\prime$, where $u = 5x$ and $v=\cos x$. $u^\prime=5$ and $v^\prime=-\sin x$. So $y^\prime=5\cos x-5x\sin x$.

Step2: Evaluate the derivative at $x = \pi$ to find the slope $m$

Substitute $x=\pi$ into $y^\prime$. $m = 5\cos\pi-5\pi\sin\pi$. Since $\cos\pi=- 1$ and $\sin\pi = 0$, then $m=5\times(-1)-5\pi\times0=-5$.

Step3: Use the point - slope form $y - y_1=m(x - x_1)$ to find $b$

The point $(x_1,y_1)=(\pi,-5\pi)$. Substitute into $y - y_1=m(x - x_1)$: $y+5\pi=-5(x - \pi)$. Expand to get $y+5\pi=-5x + 5\pi$. Then $y=-5x$, so $b = 0$.

Answer:

$m=-5$, $b = 0$