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Math Help - Finding the 79th derivative of something

  1. #1
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    Question Finding the 79th derivative of something

    Hello everyone, I am currently trying to solve a problem that invloves finding the 79th derivative of something and am trying to look for a pattern, but am not sure if I'm doing this right.

    Sin8u,

    I've gotten the following:

    cosu*8
    -64cos8x
    -512sin8x
    4096sin8x

    Any advice on this would be greatly appreciated. I want to know how to do this!
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  2. #2
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    If f(x) = \sin(ax+b), then f^{(n)}(x) = a^n\sin(ax+b+\frac{n\pi}{2}). You can prove this by induction.
    Last edited by TheCoffeeMachine; May 6th 2011 at 02:05 PM. Reason: Oops, I somehow thought that the OP wanted the 79th derivative of cos(8u). Changed the formula now.
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  3. #3
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    I would probably do it this way. Note that

    y(u)=\sin(8u)

    y'(u)=8\cos(8u)

    y''(u)=-64\sin(8u)

    y'''(u)=-512\cos(8u)

    y^{(4)}(u)=4096\cos(8u).

    Now you can see the pattern: sin, cos, -sin, -cos, and then you repeat. The coefficient just goes as the power of 8. So the question really boils down to this: what is the remainder when you divide 79 by 4?
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  4. #4
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    Okay I knew there had to be somekind of patter in this. So, I divided 79 by 4 and got a remainder of 3. So does that mean I pick the third choice in the sequence? So...

    Something like 2.20X10^(71)-cose(8U)?
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  5. #5
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    Okay I did end up getting it right and understand the whole process, thank you everyone that helped and took the effort. Thank you again!
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  6. #6
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    You should write it a bit better. Here is WolframAlpha's answer.
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  7. #7
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    Quote Originally Posted by tastybrownies View Post
    Hello everyone, I am currently trying to solve a problem that invloves finding the 79th derivative of something and am trying to look for a pattern, but am not sure if I'm doing this right.

    Sin8u,

    I've gotten the following:

    cosu*8
    -64cos8x
    -512sin8x
    4096sin8x

    Any advice on this would be greatly appreciated. I want to know how to do this!
    y = sin(8x) = Im\left[e^{8ix}\right].

    Therefore:

    \frac{d^{79}y}{dx^{79}} = \frac{d^{79}}{dx^{79}} Im\left[e^{8ix}\right] = Im\left[ \frac{d^{79} e^{8ix}}{dx^{79}}\right] = Im\left[ (8i)^{79} e^{8ix} \right]

     = -8^{79}Im\left[i e^{8ix} \right] = -8^{79} \cos(8x).
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