<s>[INST] Finding derivative of $f(x)= \frac{2x^3-4x^2+3}{x^2}$ I am trying to solve it but could not understand it. 
$f(x)= \frac{2x^3-4x^2+3}{x^2}$
 [/INST] Use the quotient rule and other basic rules
$$f'(x)=\frac{ x^2 (6x^2-8x)-(2x^3-4x^2+3)(2x)}{x^4}=\frac{6x^4-8x^3-4x^4+8x^3-6x}{x^4}=\frac{2x^4-6x}{x^4}$$$$=2-\frac{6}{x^3}$$
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