Posted by **Lisa** 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? -
**Reiny**, 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? -
**Lisa**, 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? -
**Lisa**, Sunday, March 10, 2013 at 6:06pm
*look

- Calculus Derivative- Taylor Series? -
**Reiny**, 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? -
**Anonymous**, Sunday, March 10, 2013 at 6:53pm
Thank you!

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