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example $y= 3x^3$
$y'= 9x^2$

I can solve this one but when the question come like $y=6x^4-\frac{12x^3}{3x}$ I can not solve. the same this question too $$y= \frac{x^5+3x^3-2x^2}{x}$$ and the answer is $4x^3+6x-2$ I don't know how to solve.

Can Mathematician help me?

thanks in advance.

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    but the answer is 6x^3-42011-09-29
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    ok! I understood now
    $2x^3-4x$
    $= 6x^2-4$
    2011-09-29
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    Please don't use / to denote division. It is unclear whether you mean $$\frac{x^5 + 3x^3 - 2x^2}{x}$$ or $$x^5 + 3x^3 - \frac{2x^2}{x}.$$ Same with the first problem. The only way the original question had an answer of $6x^3-4$ is if you **miscopied** it and the actual problem was $$y=\frac{6x^3-12x^2}{3x}.$$Note the square instead of the cube in the second summand.2011-09-29
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    thx! for that @ArturoMagidin I will not use next time!2011-09-29

4 Answers 4

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Notice that $$\frac{12x^3}{3x} = \frac{12}{3}\,\frac{x^3}{x} = 4x^2$$ so that you have $y=6x^4 - 4x^2$. Now use the fact that the derivative of a difference is the difference of the derivatives (if they both exist) and go from there.

If the original problem was $$y = \frac{6x^4-12x^3}{3x},$$ then $$y = \frac{6x^4}{3x} - \frac{12x^3}{3x} = 2x^3 - 4x^2;$$ given your comment, though, it seems you miscopied the problem and the $x^3$ should have been an $x^2$, i.e., $$y = \frac{6x^4 - 12x^2}{3x} = \frac{6x^4}{3x} - \frac{12x^2}{3x} = 2x^3 - 4x.$$ Then you can take derivatives. The second problem, assuming it's $$y = \frac{x^5+3x^3-2x^2}{x}$$ is solved the same way: $$y = \frac{x^5 + 3x^3 - 2x^2}{x} = \frac{x(x^4+3x^2-2x)}{x} = x^4 +3x^2 - 2x.$$

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    but the answer is $6x^3-4$2011-09-29
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    @SbSangpi: I don't know what $6x^3-4$ is an answer to, but it's **not** an answer to "the derivative of $6x^4 - \frac{12x^3}{3x}$. If by chance you actually meant $$\frac{6x^4-12x^3}{3x}$$instead, then $$\frac{6x^4-12x^3}{3x} = \frac{6x^4}{3x}-\frac{12x^3}{3x} = 2x^3 - 4x^2$$and the correct derivative is $6x^2-8x$. Not my fault if you copied the problem incorrectly.2011-09-29
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First, write $f(x) = \frac{6x^4 - 12x^3}{3x} = \frac{6x^4}{3x} - \frac{12x^3}{3x}2x^3 - 4x^2$

Then differentiate as you did in the other example. We used polynomial division to arrive at the result above.

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    can you explain more by step how do you get 2x^3-4x^22011-09-29
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    ok! I get it~ thx2011-09-29
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    I think you've misinterpreted the OPs equation, the $3 x$ only divides the second term.2011-09-29
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    @rcollyer, the question has changed. As it stands, my answer is - at worst - a fully worked example.2011-09-29
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    but the answer is 6x^3-42011-09-29
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    Well... maybe this would be a good time to learn the indefinite integral, for the purpose of reverse engineering your problem!2011-09-29
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How about you just divide $\dfrac{12x^3}{3x}$ and deal with the result? Then it's just a regular old polynomial, and you can go on term by term.

Alternately (and a much worse plan), you could wait until you learn the quotient and product rules for differentiation. But those really aren't necessary here.

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    I'll give you some voting love!2011-09-29
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Alternatively $\displaystyle\rm\ \ y\: =\: \frac f{x}\: \ \Rightarrow\ \ x\ y\: =\: f\ \ \Rightarrow\ \ x\ y' + y\: =\: f\:\:'\:\ \Rightarrow\ \ y' =\: \frac{f\:\:'-y}x\: =\: \frac{f\:\:'}x - \frac f{x^2}\:.\:$

Though this is more work than cancelling $\rm\:x\:$ from $\rm\:f\:,\:$ it works more generally - something you'll soon see when you learn how to differentiate general fractions (the quotient rule for derivatives).