How do you find the derivative of #1/(x^3-x^2)#?

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Answer 1 · 2016-04-07T12:47:52

Use either the quotient rule or the power rule (and chain) to get #-(3x^2-2x)/(x^3-x^2)^2 #

Quotient

#d/dx (1/(x^3-x^2)) = ((0)(x^3-x^2) - (1)(3x^2-2x))/(x^3-x^2)#

# = -(3x^2-2x)/(x^3-x^2)^2#

Power (and chain)

#d/dx (1/(x^3-x^2)) = d/dx ((x^3-x^2)^-1)#

# = (-1)(x^3-x^2)^-2 * d/dx(x^3-x^2)#

# = (-1)(x^3-x^2)^-2 * (3x^2-2x)#

# = -(3x^2-2x)/(x^3-x^2)^2#

A different way to write the answer

#-(3x^2-2x)/(x^3-x^2)^2 = -(x(3x-2))/(x^4(x-1)^2#

#= -(3x-2)/(x^3(x-1)^2#