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

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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

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