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    Ok so I need to find up to the sixth derivative of ln(cosx) so I can find the first 3 terms of its power series, I ended up with the power series being:

     \frac{-x^2}{2} - \frac{x^4}{12} + \frac{x^6}{45}

    However the answer has a minus in front of the final term instead of the plus I have.

    After checking wolfram alpha I found out it is because I worked out the sixth derivative to be 16 when x=0 however it is meant to be -16.

    I'm going to list my derivatives here, could someone please check if they are correct please?

     f(x)= ln(\cos x)

f'(x)= \dfrac{-\sin x}{\cos x}

f''(x)= -(\cos x)^{-2}

f'''(x)= -2\sin x(\cos x)^{-3}

f''''(x)= -2(\cos x)^{-2} + 6(\sin x)^2(\cos x)^{-4}

f'''''(x)= 16\sinx(\cos x)^{-3} + 24(\sin x)^3(\cos x)^{-5}

f''''''(x) = 16(\cos x)^{-2} -48(\sin x)^2(\cos x)^{-4} +72(\sin x)^2(\cos x)^{-4} -120(\sin x)^4(\cos x)^{-6}

    And from the sixth derivative I can see that when x=0 it will equal 16.

    Thanks, I anticipate I have missed a sign somewhere but I've been checking it for ages now and can't see anything wrong with it :/
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    (Original post by Davelittle)
    Thanks, I anticipate I have missed a sign somewhere but I've been checking it for ages now and can't see anything wrong with it :/
    To make life easier, note that:
     f'(x) = -\tan(x)
     f''(x) = -\sec^2(x)
     f^3(x) = -2\sin(x)(\cos(x))^{-3} = -2\tan(x)\sec^2(x)  = -2f'(x)f''(x)

    So
     f^4(x) = -2(f''(x))^2  - 2f'(x)f^3(x)
     f^5(x) = -4f''(x)f^3(x) - 2f''(x)f^3(x) - 2f'(x)f^4(x)
     f^6(x) = -4(f^3(x))^2 - 8f''(x)f^4(x) - 2(f^3(x))^2 - 2f'(x)f^5(x)

    (Sorry, notation is a bit rubbish...)

    Evaluating the odd derivatives yields 0 so we only need to worry about the even ones:
     f''(0) = -1
     f^4(0) = -2
     f^6(0) = -16

    You had a sign error on  f^4(x) . it should be  -2(\cos(x))^{-2} - 6sin^2(x)(\cos(x))^{-4}
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    (Original post by Blazy)
    To make life easier, note that:
     f'(x) = -\tan(x)
     f''(x) = -\sec^2(x)
     f^3(x) = -2\sin(x)(\cos(x))^{-3} = -2\tan(x)\sec^2(x)  = -2f'(x)f''(x)

    So
     f^4(x) = -2(f''(x))^2  - 2f'(x)f^3(x)
     f^5(x) = -4f''(x)f^3(x) - 2f''(x)f^3(x) - 2f'(x)f^4(x)
     f^6(x) = -4(f^3(x))^2 - 8f''(x)f^4(x) - 2(f^3(x))^2 - 2f'(x)f^5(x)

    (Sorry, notation is a bit rubbish...)

    Evaluating the odd derivatives yields 0 so we only need to worry about the even ones:
     f''(0) = -1
     f^4(0) = -2
     f^6(0) = -16

    You had a sign error on  f^4(x) . it should be  -2(\cos(x))^{-2} - 6sin^2(x)(\cos(x))^{-4}
    Thanks so much, messed up my integration of  \cos x after all that work -.-

    Appreciate it!
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    (Original post by Blazy)
    To make life easier, note that:
     f'(x) = -\tan(x)
     f''(x) = -\sec^2(x)
     f^3(x) = -2\sin(x)(\cos(x))^{-3} = -2\tan(x)\sec^2(x)  = -2f'(x)f''(x)

    So
     f^4(x) = -2(f''(x))^2  - 2f'(x)f^3(x)
     f^5(x) = -4f''(x)f^3(x) - 2f''(x)f^3(x) - 2f'(x)f^4(x)
     f^6(x) = -4(f^3(x))^2 - 8f''(x)f^4(x) - 2(f^3(x))^2 - 2f'(x)f^5(x)

    (Sorry, notation is a bit rubbish...)

    Evaluating the odd derivatives yields 0 so we only need to worry about the even ones:
     f''(0) = -1
     f^4(0) = -2
     f^6(0) = -16

    You had a sign error on  f^4(x) . it should be  -2(\cos(x))^{-2} - 6sin^2(x)(\cos(x))^{-4}
    Carrying on from here only gives -8 for the sixth derivative, have I also missed a factor of 2 somewhere?
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    (Original post by Davelittle)
    Carrying on from here only gives -8 for the sixth derivative, have I also missed a factor of 2 somewhere?
    How are you going about doing it? Once you fixed the minus sign on your fourth derivative, then it should just be a case of fixing the signs...

    For example,  f^5(x) should be the negative of what you have in the first post.


    Your f^5(x) isn't right... - please show your workings.


    If you're using my method, then you just need to make sure you're plugging the right numbers.
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    (Original post by Blazy)
    How are you going about doing it? Once you fixed the minus sign on your fourth derivative, then it should just be a case of fixing the signs...

    For example,  f^5(x) should be the negative of what you have in the first post.


    Your f^5(x) isn't right... - please show your workings.


    If you're using my method, then you just need to make sure you're plugging the right numbers.
    No worries, managed to spot the error!

    Thanks again, this whole repeated differentiation seems to lead me to sign errors quite often!
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    (Original post by Davelittle)
    No worries, managed to spot the error!

    Thanks again, this whole repeated differentiation seems to lead me to sign errors quite often!
    It's tedious, but you just need to practice more and stamp out the errors!
 
 
 
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