anonymous
  • anonymous
How do you solve for higher order derivatives for example 90th derivative of cos(2x)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
http://www.youtube.com/watch?v=wV1FrqwZyKw&feature=feedlik
anonymous
  • anonymous
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)
anonymous
  • anonymous
thanks so much

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