Find the derivative of y=1/(x^3+3x^2+2x)?

Since the function is a ratio, we'll apply the quotient
rule to evaluate it's first derivative:

(u/v) = (u'*v -
u*v')/v^2 (*)

We'll put u = 1 => u' =
0

v = x^3+3x^2+2x => v' = 3x^2 + 6x +
2

We'll substitute u,v,u',v' into the formula
(*):

f'(x) = [0*(x^3+3x^2+2x) - 1*(3x^2 + 6x +
2)]/(x^3+3x^2+2x)^2

f'(x) = -(3x^2 + 6x +
2)/(x^3+3x^2+2x)^2

The first derivative of
the given function is:

f'(x)
= -(3x^2 + 6x +
2)/(x^3+3x^2+2x)^2

or, written
in a simplifed form

f'(x) =
-(3x^2 + 6x + 2)/x^2*(x+1)^2*(x+2)^2