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

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

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

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

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

Did I tackle this question correctly?

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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? –  ae0709 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: <tex.stackexchange.com/q/14821>; –  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

1 Answer 1

up vote 4 down vote accepted

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

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

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

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

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

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OH i got it its - on the bottom oh man -____- close enough haha –  soniccool Oct 25 '11 at 0:57
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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
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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

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