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Math Help - Another Derivative of a Sin Function Problem

  1. #1
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    Another Derivative of a Sin Function Problem

    Let f(x)= 9sin(x+π) + cos2x

    a) Find f'(x)
    b) Find f''(x)
    c) Find f'''(x)

    Would the first derivative be this equation?

    f'(x)= 9 * Sin(x+π)*1 + 0*Sin(x+π) + -2sin(2x)?

    Would that be correct, using the chain rule, etc?

    Thanks!
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  2. #2
    Super Member Deadstar's Avatar
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    Almost but you're forgetting to differentiate the 'sin' part of sin(x + n) so the answer should be...
    9\cos(x+n) - 2\sin(2x)

    Is it an n or a pi? Won't matter anyway answer would be the same but exchange the n for a pi!
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  3. #3
    MHF Contributor Calculus26's Avatar
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    sin(x+ pi) ' = cos(x+ pi)


    d/dx(9sin(x+ pi) +cos(2x)) = 9cos(x+ pi) - 2sin(2x)
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  4. #4
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    Ok so from there, let's see if the 2nd derivative is correct that I have.

    f''(x)= 9cos(x+pi) - 2sin(2x)
    f''(x)= 9(-sin(x+pi)) - [2cos(2x)*2]
    f''(x)= -9sin(x+pi) - 4cos(2x)

    So would the second derivative be:

    f''(x)= -9sin(x+pi) - 4cos(2x)


    Thanks if anyone could verify that.
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  5. #5
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    Quote Originally Posted by r2d2 View Post
    Ok so from there, let's see if the 2nd derivative is correct that I have.

    f''(x)= 9cos(x+pi) - 2sin(2x)
    f''(x)= 9(-sin(x+pi)) - [2cos(2x)*2]
    f''(x)= -9sin(x+pi) - 4cos(2x)

    So would the second derivative be:

    f''(x)= -9sin(x+pi) - 4cos(2x)


    Thanks if anyone could verify that.
    that is correct.
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  6. #6
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    And finally, the 3rd Derivative!

    f'''(x)= -9sin(x+pi) - 4cos2x
    f''''(x)= -9(cos(x+pi)) - [4sin(2x)*-2]
    f'''(x)= -9cos(x+pi) - [-8sin(2x)]

    f'''(x)= -9 cos(x+pi) + 8sin2x

    This would be the 3rd derivative, correct?

    Thanks will be going around for all!
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  7. #7
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    Quote Originally Posted by r2d2 View Post
    And finally, the 3rd Derivative!

    f'''(x)= -9sin(x+pi) - 4cos2x
    f''''(x)= -9(cos(x+pi)) - [4sin(2x)*-2]
    f'''(x)= -9cos(x+pi) - [-8sin(2x)]

    f'''(x)= -9 cos(x+pi) + 8sin2x

    This would be the 3rd derivative, correct?

    Thanks will be going around for all!
    will you require the same reassurance when you take the 99th derivative?

    have some confidence in your calculations.
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  8. #8
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    Well my BC teacher didn't assign the 99th derivative but if he did, I probably would be checking myself until then!

    I definitely understand this derivative of trigonometric functions now. Thanks a lot.
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