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For a homework assignment, I was asked to find the derivative of the function $$f(x)= \frac{x^{3}-6x^{2}+6}{x^{2}}.$$

Using the product rule, I worked through the following: $$f'(x)=\frac{\frac{d}{dx}(x^{3}-6x^{2}+6)}{x^{2}}+(x^{3}-6x^{2}+6)\left[\frac{d}{dx}\frac{1}{x^{2}}\right]$$

Since the derivative of $\frac{1}{x{^2}}$ is $-\frac{2}{x^{3}}$, the left side of the formula resulted in $$\frac{-2(x^{3}-6x^{2}+6)}{x^{3}}= -\frac{2x^{3}+12x^{2}-12}{x^{3}}.$$

The right side of the equation split up by the product rule into $$\frac{\frac{d}{dx}(x^{3})-6(\frac{d}{dx}(x^{2}))+\frac{d}{dx}(6)}{x^{2}}$$

The derivatives of $x^{3}$ is $3x^{2}$; $x^{2}$ is $2x$, and 6 is 0, thus making the formula \ $$\frac{3x^{2}+12x}{x^{2}}-\frac{2x^{3}-12x{^2}+12}{x^{3}}$$

That's where I got stuck. Now, when I plugged the formula into WolframAlpha, the derivative calculated was $1-\frac{12}{x^{3}}$ How can I go from the formula above to the calculated result?

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It may help to notice $f(x)$ can be written as $f(x) = x - 6 + \frac{6}{x^2}$. As far as combining the fractions, just get a common denominator. I will also point out (though not exactly where, that's for you to find) that you switched a negative sign somewhere in your calculation. – JavaMan Feb 15 '11 at 3:04
You dropped a sign when dealing with the 1/(x^2) part. The +12x^2 in the numerator should be -12x^2 due to the leading minus sign. Then those terms will cancel at the end. – Ross Millikan Feb 15 '11 at 3:07
Note that you got all the Leibnitz notation derivatives wrong. None of them are $\frac{d}{dy}$, they should all be $\frac{d}{dx}$; you are taking derivatives with respect to $x$, not with respect to $y$. – Arturo Magidin Feb 15 '11 at 3:12
up vote 3 down vote accepted

As noted in other answers, you have a +/- transcription error in the numerator of the first term of your answer—it should be:

$$\begin{align} \frac{3x^{2}-12x}{x^{2}}-\frac{2x^{3}-12x{^2}+12}{x^{3}} &=\frac{(3x^{2}-12x)\cdot x}{(x^{2})\cdot x}-\frac{2x^{3}-12x{^2}+12}{x^{3}} \\ &=\frac{3x^{3}-12x^2}{x^{3}}-\frac{2x^{3}-12x{^2}+12}{x^{3}} \\ &=\frac{3x^{3}-12x^2-(2x^{3}-12x{^2}+12)}{x^{3}} \\ &=\frac{3x^{3}-12x^2-2x^{3}+12x{^2}-12}{x^{3}} \\ &=\frac{x^{3}-12}{x^{3}} \\ &=\frac{x^{3}}{x^3}-\frac{12}{x^{3}} \\ &=1-\frac{12}{x^{3}} \end{align}$$

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It was multiplying the left formula by x that made me follow the process through, and thus understand why the WA answer came about – Jason Feb 15 '11 at 18:12

Are you specifically told to use the product rule with no simplication? If not, you should probably try dividing through by $x^2$ first to simplify what you're doing.

Notice $$ f(x)=\frac{x^3-6x^2+6}{x^2}=x-6+6x^{-2}. $$ This should be much easier to work with and gives you the derivative given by WolframAlpha as well.

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OK. I am confused so i am working the solution out again completely. What i feel is that somewhere you have a made a mistake of $\pm$ sign. That's all not a major error.

You have $$\Bigl[x^{3}-6x^{2}+6\Bigr] \cdot -\frac{2}{x^{3}} \quad \ (1)$$ and $$\Bigl[3x^{2}-12x\Bigr] \cdot \frac{1}{x^{2}}\quad \ (2)$$

So you have to add these two. Eq (1) is : $ -2 + \frac{12}{x} -\frac{12}{x^{3}}$ and Eq (2) is: $3 - \frac{12}{x}$. Adding these two gives $1- \frac{12}{x^{3}}$ which is what you wanted.

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I would recommend writing $a(-b)$ or $a\cdot(-b)$ if you want to denote the product of $a$ and $-b$ in that order, rather than $a\cdot-b$, because the latter can too easily be interpreted as the difference $a-b$. – Jonas Meyer Feb 15 '11 at 3:49

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