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Posted by on Sunday, March 10, 2013 at 5:37pm.

let f(x)= x/x-1

find f'(x) f ''(x) and a formula for f ^ (n) * x.

I found the first and second derivatives but do not know how to make a general equation for this. I have not learnt the Taylor or Maclaurin Series either. Thank you.

  • Calculus Derivative- Taylor Series? - , Sunday, March 10, 2013 at 5:51pm

    your first derivative should have been:
    y' = -1(x-1)^-2
    then y'' = 2(x-1)^-3
    y''' = -6(x-1)^-4 ----> the third derivativ
    y'''' = 24(x-1)^-5 ----> the 4th derivative
    y''''' = -120(x-1)^-6 ---> the 5th derivative

    did you notice that the numbers 1 , 2, 6, 24 , 120 are factorials ?
    Did you notice that the alternate ±
    did you notice that the exponent is -(n+1) for the nth derivative, e.g. , for the 4th derivative the exponent is -5

    how about a general derivative expression from the above?

  • Calculus Derivative- Taylor Series? - , Sunday, March 10, 2013 at 6:06pm

    Yes I did get those as my derivatives. Thank you very much for your help Reiny.

    I did not know which patterns to look for.
    I think that it would be (-1)^n for the alternating neg. and pos.?

    Then n! for the factorial? And lastly x^ -(n+1) for the power of x in the denominator?

    I got: f^ (n) (x)= (-1)^n * n! * x^ -(n+1) Is this right?

    Thank you very much again. I never knew what patterns to help for. Now I know how to do such problems :)

  • Calculus Derivative- Taylor Series? - , Sunday, March 10, 2013 at 6:06pm

    *look

  • Calculus Derivative- Taylor Series? - , Sunday, March 10, 2013 at 6:27pm

    good job

    especially the (-1)^n part, good of you to notice that if n is even , the result has to be positive, and if n is odd we need a negative.
    (-1)^n will do that

  • Calculus Derivative- Taylor Series? - , Sunday, March 10, 2013 at 6:53pm

    Thank you!

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