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Math Help - Finding the third derivative

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    Finding the third derivative

    I really don't like trig functions so could someone give this the once over?

    "Find f'''(x) of f(x) 6x^5 + 3xcosx"

    f'(x) = 30 x^4 + 3x(-sinx) + 3cosx

    f''(x) = 120x^3 + -3x(cosx) + -sinx

    f'''(x) = 360x^2 + 3x(sinx) + -cosx

    Did I do this right???

    Thanks in advance
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    Quote Originally Posted by calcbeg View Post
    I really don't like trig functions so could someone give this the once over?

    "Find f'''(x) of f(x) 6x^5 + 3xcosx"

    f'(x) = 30 x^4 + 3x(-sinx) + 3cosx
    Looking good up to this point!

    f''(x) = 120x^3 + -3x(cosx) + -sinx
    I'm not sure where the last term came from, and you didn't handle the product rule correctly here.

    f'(x) = 30x^4-3x\sin x + 3\cos x

    So

    f''(x) = 120x^3-3(x\cos x + \sin x) - 3\sin x

    = 120x^3-3x\cos x-3\sin x-3\sin x

    = 120x^3-3x\cos x-6\sin x

    and

    f'''(x) = 360x^2-3(-x\sin x + \cos x)-6\cos x
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