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Finding a higher order derivative of a trig function

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the 73rd derivative of f(x) = sin(2x) + 3
    (Hint: Take the first five derivatives to find a pattern)

    2. Relevant equations

    dy/dx

    3. The attempt at a solution
    I took the first five derivatives to find the pattern:
    dy/dx = 2cos2x
    d2y/dx2 = -4sin2x
    d3y/dx3 = -8cos2x
    d4y/dx4 = 16sin2x
    d5y/dx5 = 32cos2x

    Now if we look at the pattern we can see that every odd term derivative will result in tcos2x and every even term derivative will result in tsin2x where t is an element of all integers.
    If we just look at the value of t and not the sign, let's say we are focusing on |t|, then we just look at the pattern and figure out that it will be 2^73 because each of the terms are multiplied by 2^n where n in this case is 73. Now if we were looking at the sign we notice that it alternates: (every underlined term number will be a negative derivative, otherwise it will be a postive derivative)
    1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10
    11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20
    21, 22 , 23
    31, 32 , 33
    41, 42 , 43
    51, 52 , 53
    61, 62 , 63
    71, 72 , 73

    A pattern starts after the 20th term because then the cycle repeats and by this we know that the 73rd derivative will be a positive derivative so a positive t, thus the 73rd derivative is 2^73cos2x.
    I am fairly sure that my answer is correct, however i do not think that my process and explanation are sufficient enough mathematically to show my answer. Can you please help in finding a general formula for this or perhaps a more mathematical way of finding the derivative, especially for the part where i was determining the sign (positive,negative) of the derivative. Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2011 #2
    I don't think the sign pattern should have changed at the 20th derivative. It should remain a cycle of four.
     
  4. Oct 30, 2011 #3

    lurflurf

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    sin(n)(x)=sin(x+n π/2)
    (f(ax+b))(n)=anf(n)(ax+b)

    and write 73=4*18+1
     
  5. Nov 1, 2011 #4
    I am still confused about what you mean, is this the general formula to finding the derivative of this function to any number and you plug in 73 for n? Also sin(x+n π/2)? is it n * n/2 or n + n/2? Please go over your post in more detail because i do not understand how to apply any of the equations you wrote or what they mean. Thank you.
     
  6. Nov 1, 2011 #5
    tmlrlz: It looks like you're on the right track. I agree with the 2^73 part. How did you get that the pattern changes after the 20th derivative? It would certainly seem odd to me if the pattern just changes for no apparent reason after the 20th derivative. I mean, the sign changes happen because of the way the derivatives of sin and cos work.
     
  7. Nov 1, 2011 #6
    When i was writing down the terms i thought that it repeated after 20 terms but someone pointed out that it repeats after four terms. however i just dont know how to make this answer seem more mathematical, like a formula for the derivative that i could plug in 73 and get out the 73rd derivative because the alternating cos and sin as well as the pattern of positives an negatives
     
  8. Nov 1, 2011 #7
    i think there are 3 parts, like you mentioned: the sign, whether it's sin or cos, and what the number in front of the sin/cos is (disregarding the sign). You've already figured out the number (2^73). The pattern for sin/cos is: cos, sin, cos, sin, cos, sin... And I think you found the pattern for the sign too. Does that help?
     
  9. Nov 1, 2011 #8

    lurflurf

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    Yes that is the general equation for the nth derivative of sine the second thing is pi, not another n.
    [tex]\sin^{(n)}(x)=\frac{d^n}{dx^n}\sin(x)=\sin(x+\frac{\pi}{2}n)[/tex]

    The second thing is the chain rule
    [tex]\frac{d^n}{dx^n}f(a x)=a^n f^{(n)}(a x)[/tex]

    Your question can be answered by combining these two. Also as has been mentioned there is a repeating pattern, 73=18*4+1 so the 73 derivative is a repeat of earlier ones.
     
  10. Nov 1, 2011 #9
    I dont understand how you would combine these two to get a formula for the derivative. If you plug in 73 it doesn't turn out to be the same thing and how does it show that sin and cos alternate every other term and the pattern of the positives and negatives? I'm sorry, i'm sure that what you're saying is correct, i just do not understand. Thank you.
     
  11. Nov 1, 2011 #10

    lurflurf

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    Homework Helper

    combine them like
    [tex]\frac{d^n}{dx^n}\sin(a x)=a^n \sin^{(n)}(a x)=a^n \sin(a x+\frac{\pi}{2}n)[/tex]
    We can also eliminate the shift if desired
    [tex]\sin^{(101)}(x)=\frac{d^{101}}{dx^{101}}\sin(x)= \sin(x+\frac{\pi}{2}101)= \sin(x+25*2\pi+\frac{\pi}{2})=\sin(x+\frac{\pi}{2})=\cos(x)[/tex]
    [tex]\sin^{(1003)}(x)=\frac{d^{1003}}{dx^{1003}}\sin(x)=\sin(x+\frac{\pi}{2}1003)= \sin(x+250*2\pi+3\frac{ \pi }{2})=\sin(x+3\frac{\pi}{2})=-\cos(x)[/tex]
     
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