# Did I tackle this implicit differentiation problem correctly?

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

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

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

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

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

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? – 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: <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

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

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

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

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

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

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