What is the derivative of $1/(x^3-x^2)$ ?

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Answer 1 · 2015-05-29T19:41:00

One exponential rule states that $1/a^n=a^-n$ , thus, we can rewrite the expression as

$(x^3-x^2)^-1$

Now, we can rename $u=x^3-x^2$ and use the chain rule to derivate it, as follows:

$(dy)/(dx)=(dy)/(du)(du)/(dx)$

$(dy)/(du)=-u^-2$

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

Now, $(dy)/(dx)=-u^-2(3x^2-2x)$

Now, substituting $u$ :

$(dy)/(dx)=-(x^3-x^2)^-2(2x^2-2x)$

$(dy)/(dx)=-(2(x^2-x))/(x^3-x^2)^2$

What is the derivative of 1/(x^3-x^2)1/(x^3-x^2)?