can someone help me determine derivatives?

Question

What is the first and second derivative of this?

f(x)= x/(x-1)^2

Answer 1

take f[x] = x / g[x] = (x-1)^2

Quotient rule states that f'[x]g[x] - f[x]g'[x]/ (g[x])^2
f[x]= x
f'[x]= 1
g[x]= (x-1)^2
g'[x]= 2(x-1)

Plug in those values to get the first derivative and do the same with the 2nd derivative.

Answer 2

Use quotient and chain rules to find the first derivative:
f'(x) = [ 1*(x-1)^2 - x*2(x-1)*(1) ] / ((x-1)^2)^2
f'(x) = -(x+1) / (x-1)^3

Take the derivative of the first derivative to get the 2nd derivative:
f''(x) = [-1 * (x-1)^3 - ((3*(x-1)^2*1) * -(x+1) ] / ((x-1)^3)^2
f''(x) = 2(x+2) / (x-1)^4

Answer 3

use quotient rule.
f(x)=u(x)/v(x)
f'(x)=[u(x)v'(x)-v(x)u'(x)]/v^2(x)
u(x)=x
v(x)=(x-1)^2
dy/dx=[x (2)(x-1) + (x-1)^2(1)] / (x-1)^4
=[2x^2-2x+x^2-2x+1]/(x-1)^4
=[3x^2-4x+1]/(x-1)^4
for the second derivative
u(x)=3x^2-4x+1
v(x)=(x-1)^4
use the quotient rule again
f''(x)=[(3x^2-4x+1)(4)(x-1) +(x-1)^4(6x-4)]/(x-1)^8
simplify the numerator further if necessary.