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#