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Math Help - derivative

  1. #1
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    derivative

    An equation of the tangent line to the curve:

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

    at the point is y =
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by samtheman17 View Post
    An equation of the tangent line to the curve:

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

    at the point is y =

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

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

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

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

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

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

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

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

    you can continue
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  3. #3
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    thank you!!

    how do you now solve so it looks like

    y= mx +b??
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by samtheman17 View Post
    thank you!!

    how do you now solve so it looks like

    y= mx +b??

    it is the same just simplify

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

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

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

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