1. ## derivative

An equation of the tangent line to the curve:

y= f(x) = x(9cosx - 3sinx)

at the point $(3 \pi,f(3 \pi)$ is y =

2. Originally Posted by samtheman17
An equation of the tangent line to the curve:

y= f(x) = x(9cosx - 3sinx)

at the point $(3 \pi,f(3 \pi)$ is y =

$\displaystyle f(x) = x(9\cos x - 3\sin x )$

$\displaystyle f(3\pi) = 3\pi(9\cos (3\pi) - 3 \sin (3\pi) )$

$\displaystyle f(3\pi) = 3\pi(-9) = -27\pi$

$\displaystyle f'(x) = (9\cos x - 3 \sin x ) + x ( 9(-\sin x ) - 3 \cos x )$

$\displaystyle f'(x) = (9\cos x - 3 \sin x ) + x (-9\sin x - 3 \cos x )$

$\displaystyle f'(3\pi) = (9\cos (3\pi) - 3 \sin (3\pi) ) + 3\pi (-9\sin (3\pi) - 3 \cos (3\pi) )$

$\displaystyle f'(3\pi) = -9 + 9\pi$

$\displaystyle \frac{y-(-27\pi)}{x-3\pi} = -9 + 9\pi$

you can continue

3. thank you!!

how do you now solve so it looks like

y= mx +b??

4. Originally Posted by samtheman17
thank you!!

how do you now solve so it looks like

y= mx +b??

it is the same just simplify

$\displaystyle \frac{y+27\pi}{x-3\pi} = -9+9\pi$

$\displaystyle y +27\pi = (-9+9\pi)(x-3\pi)$

$\displaystyle y+27\pi = -9x +27\pi + 9\pi x - 27\pi ^2$

$\displaystyle y = - 9\pi x - 27\pi ^2+9\pi$