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anonymous

  • 5 years ago

How do you solve for higher order derivatives for example 90th derivative of cos(2x)

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  1. anonymous
    • 5 years ago
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    http://www.youtube.com/watch?v=wV1FrqwZyKw&feature=feedlik

  2. anonymous
    • 5 years ago
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    Write out a couple of the first few derivatives: f(x) = cos(2x) f'(x) = -2sin(2x) f''(x) = -4cos(2x) f'''(x) = 8sin(2x) f(4)(x) = 16cos(2x) the key is to notice the pattern. Every 4 derivatives taken the expression reverts back to cosine. With every derivative the coefficient doubles. So the 88th derivative is a positive cosine, then the 90th must be a negative cosine, with the coefficient equal to 2^90 f(90)(x) = -(2^90)cos(2x)

  3. anonymous
    • 5 years ago
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    thanks so much

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