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

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Answer 1 · 2015-06-02T15:19:57

One law of exponents states that #a^-n=1/a^n#

Thus, we can rewrite this whole expression as #f(x)=(x^3-x^2)^-1#

Now, using the chain rule, we cna rename #u=x^3-x^2#, thus making #f(x)=u^-1#

The chain rule states that

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

So,

#(dy)/(du)=-1*u^-2#

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

#(dy)/(dx)=-u^-2(3x^2-2x)#

Substituting #u#:

#(dy)/(dx)=-(x^3-x^2)^-2(3x^2-2x)=-(3x^2-2x)/(x^3-x^2)^2#=#(3x^2-2x)/(x^6-2x^5+x^4)=(cancelx(3x-2))/(cancelx(x^5-2x^4+x^3))#

Thus, final answer:

#(3x-2)/(x^5-2x^4+x^3)#