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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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    thx! for that @ArturoMagidin I will not u$s$e nex$t$ 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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    @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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    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).