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?
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.
5 answers
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 :)
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 :)
*look
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
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
Thank you!