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I am trying to solve it but could not understand it.

$f(x)= \frac{2x^3-4x^2+3}{x^2}$

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    $\begingroup$ What is the derivative of $x^n$ ? Then remember that the derivative of a sum is the sum of the derivatives. $\endgroup$ – Claude Leibovici Feb 17 '16 at 5:56
  • $\begingroup$ To add on to Claude's comment: It might be helpful to write the fraction as a sum of fractions and simplify these. $\endgroup$ – Roland Feb 17 '16 at 6:21
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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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**

HINT

** It's pretty easy. Whenever you get an expression at this level, try to simplify it as much as possible before starting. We can easily split this thing up into $$\frac {2x^3}{x^2}-\frac {4x^2}{x^2}+\frac {3}{x^2}$$. This simplifies to $$2x-4+\frac {3}{x^2}$$ Can you use power rule to solve this?

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  • $\begingroup$ Thanks guys, that makes sense!! My professor is horrible!! $\endgroup$ – Shawn Ahmed Feb 17 '16 at 6:24
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Sometimes the algebra with the quotient rule is tedious. Consider the following method if this is a problem:

$$\frac{2x^3-4x^2+3}{x^2} = 2x - 4 + \frac {3}{x^2}$$ It's fine to factor out the $x^2$ since our new expression isn't defined at $x=0$ anyway.

Using this, we get $$f'(x) = 2 + (3)(-2)(x^{-3}) = 2 - \frac{6}{x^3}$$

For the last step, I used the power rule, which I will assume you are acquainted with.

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