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.
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?
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 :)
*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
Thank you!
To find the derivatives of f(x) = x/(x-1), we can use the quotient rule.
1. First derivative (f'(x)):
Apply the quotient rule:
f'(x) = (x-1)(1) - (x)(1) / (x-1)^2
Simplify the expression:
f'(x) = (x - 1 - x) / (x-1)^2
= -1 / (x-1)^2
So the first derivative of f(x) is f'(x) = -1/(x-1)^2.
2. Second derivative (f''(x)):
To find the second derivative, differentiate the first derivative we obtained:
f''(x) = d/dx [-1 / (x-1)^2]
To simplify this, we can rewrite -1 as -1*(x-1)^(-2):
f''(x) = d/dx - (x-1)^(-2)
Apply the chain rule:
f''(x) = 2(x-1)^(-3) * d/dx (x-1)
Simplify:
f''(x) = -2(x-1)^(-3)
So the second derivative of f(x) is f''(x) = -2/(x-1)^3.
3. n-th derivative (f^(n)(x)):
To find the general formula for the n-th derivative of f(x), we can observe that the pattern in the derivatives continues. The first derivative has -1/(x-1)^2, and the second derivative has -2/(x-1)^3.
We can generalize this pattern and notice that every derivative has a factor of -n/(x-1)^(n+1).
Therefore, the general formula for the n-th derivative of f(x) is:
f^(n)(x) = (-1)^n * n / (x-1)^(n+1)
This formula holds for any positive integer value of n.
Note: If you are not familiar with Taylor or Maclaurin Series, this formula should suffice to find the n-th derivative of f(x).