Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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
add comment

4 Answers 4

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: $$y = \frac{x^5 + 3x^3 - 2x^2}{x} = \frac{x(x^4+3x^2-2x)}{x} = x^4 +3x^2 - 2x.$$

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
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
1  
I think you've misinterpreted the OPs equation, the $3 x$ only divides the second term. –  rcollyer Sep 29 '11 at 4:10
1  
@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
show 1 more comment

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.

share|improve this answer
    
I'll give you some voting love! –  The Chaz 2.0 Sep 29 '11 at 4:38
add comment

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.