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

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### 3 Answers

We have to find the derivative of y= 1/(x^3+3x^2+2x)

y = 1/(x^3+3x^2+2x) = (x^3+3x^2+2x)^-1

Using the chain rule

y' = -(x^3+3x^2+2x)^-2 * ( 3x^2 + 6x + 2)

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

**The required derivative of 1/(x^3+3x^2+2x) is -( 3x^2 + 6x + 2) / (x^3+3x^2+2x)^2**

To find the derivative of y = 1/(x^3+3x^2+2x).

We know that if y = 1/f(x), then dy/dx = y'= {-1/f(x)^2} f'(x).

Therefore y ' = {-1/(x^3+3x^2+2x)^2}(x^3+3x^2+2x).

y' = {-1/(x^3+3x^2+2x)^2}{3x^2+3*2x+2)

y' = {-1/(x^2 (x^2+3x+2)^2}(3x^2+6x+2)

y' = -(3x^2+6x+2)/{x^2(x^2+3x+2)^2}.

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**