Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$2x^3 + x^2y-xy^3 = 2$$

$$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$$

$$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y6^2 \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$$

$$\frac{\mathrm{d}y}{\mathrm{d}x} (x^2 +3y^2)(6x^2+2x-y^3) = 0$$

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6x^2+2x-y^3}{x^2+3y^2} $$

Did I tackle this question correctly?

share|cite|improve this question
I believe your $3y6^2$ should read $3xy^2$, so you are missing an $x$ in the next line (as well as some other things). – AMPerrine Oct 25 '11 at 0:56
@Srivatsan Narayanan: out of curiosity, why do you and others always remove the italicization of the differential? – mathmath8128 Oct 25 '11 at 1:59
@aengle Well, I don't use either style (italics or text) consistently. I think the idea of using text style is that $\mathrm d$ is not really a variable, but more like an operator of sorts. This is in contrast to the variable $x$ sitting right next to it. – Srivatsan Oct 25 '11 at 2:21
@SrivatsanNarayanan Here's an answer to how differentials should technically be represented: <>; – Samuel Tan Oct 25 '11 at 5:46
@aengle: there's a mention of this custom in The Not So Short Introduction to $\LaTeX$ $2\varepsilon$. (See page 69.) The recommendation goes all the way back to Knuth (but I can't seem to find where he talked about this). – J. M. Oct 25 '11 at 15:46
up vote 4 down vote accepted

$2x^3 + x^2y-xy^3 = 2$

$$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$$

$$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y^2x \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$$

$$\frac{\mathrm{d}y}{\mathrm{d}x} (x^2 - 3y^2x) + (6x^2+2xy-y^3) = 0$$

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6x^2+2xy-y^3}{3y^2x -x^2} $$

share|cite|improve this answer
OH i got it its - on the bottom oh man -____- close enough haha – soniccool Oct 25 '11 at 0:57
MY REP IS 1337. Nobody vote on anything of mine ever again! – The Chaz 2.0 Oct 25 '11 at 1:03
LOL I WAS ABOUT TOO THO! – soniccool Oct 25 '11 at 1:08
Wait wait how did you get the 3y^2x-x2 where did that x come from? – soniccool Oct 25 '11 at 1:09
That's what I was referring to in my comment. Product rule on $xy^3$ gives $1y^3+3y^2x\frac{dy}{dx}$. – AMPerrine Oct 25 '11 at 1:12

Your Answer


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.