How do i find the derivative of g(x)=x/(x-1)^2?

Question

and what are the critical numbers?

Answer 1

Using Quotient Rule:-
g `(x) = [(x - 1)² (1) - (x) (2)(x - 1) ] / (x - 1)^4
g `(x) = [ (x - 1)[(x - 1) - 2x] ] / (x - 1)^4
g `(x) = [ (x - 1)(- 1 - x)] / (x - 1)^4
g `(x) = - (x + 1) / (x - 1)³

Using Product Rule:-
g(x) = (x) (x - 1)^(-2)
g `(x) = (1)(x - 1)^(-2) + (-2)(x - 1)^(-3)(x)
g `(x) = (x - 1)^(-3) [ (x - 1) - 2x ]
g `(x) = - (x + 1) / (x - 1)³

Answer 2

Use the quotient rule, or you could also use the product rule by writing it in the form:

g(x) = x (x-1)^-2

Ill use the quotient rule:

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

The quotient rule states that:

If g(x) = u(x) / v(x), then g'(x) = [v(x)u'(x) - u(x)v'(x)] / [v(x)]^2

Now just apply that, where u(x) = x and v(x) = (x-1)^2

g'(x)
= [(x-1)^2 - 2x(x-1)] / (x-1)^4
= [x^2 + 1 - 2x - 2x^2 + 2x] / (x-1)^4
= (1 - x^2) / (x-1)^4
= (1-x)(1+x) / (x-1)^4
= -(x-1)(x+1) / (x-1)^4
= -(x+1) / (x-1)^3

To find the critical points, set the derivative equal to zero and solve. This will give the x coordinates of all stationary points. To find the coordinates of the inflection points, find where the second derivative is equal to zero.