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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^3-4 – Sb Sangpi Sep 29 '11 at 4:15
ok! I understood now <br> $2x^3-4x$ <br> $= 6x^2-4$ – Sb Sangpi Sep 29 '11 at 4:26
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. – Arturo Magidin Sep 29 '11 at 4:28
thx! for that @ArturoMagidin I will not use next time! – Sb Sangpi Sep 29 '11 at 4:33
up vote 3 down vote accepted

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 $6x^3-4$ – Sb Sangpi Sep 29 '11 at 4:21
@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. – Arturo Magidin Sep 29 '11 at 4:26

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^2 – Sb Sangpi Sep 29 '11 at 4:08
ok! I get it~ thx – Sb Sangpi Sep 29 '11 at 4:09
I think you've misinterpreted the OPs equation, the $3 x$ only divides the second term. – rcollyer Sep 29 '11 at 4:10
@rcollyer, the question has changed. As it stands, my answer is - at worst - a fully worked example. – The Chaz 2.0 Sep 29 '11 at 4:16
but the answer is 6x^3-4 – Sb Sangpi Sep 29 '11 at 4:17

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! – The Chaz 2.0 Sep 29 '11 at 4:38

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).

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