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$$4x^3 + x^2y - xy^3 = 4$$

This is what I have so far:

$$(2xy + x^2 y') - (y^3 + 3xy^2 y') = -12x^2$$

Should I bring everything but the y primes over to the right side by dividing it? I'm not so sure on what to do in this situation.

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Latex! Try it, you'll like it! And so will we! – Robert Lewis Oct 13 '13 at 4:41
I've edited your question to include MathJax. Please verify that it means what you intend. – user61527 Oct 13 '13 at 4:42
@RobertLewis I'm not sure if that's sarcasm or not, but I wasn't aware that you could use latex in a post. Yes,@T.Bongers, that is what I was trying to convey. – igknighton Oct 13 '13 at 4:45
Sarcasm was not my intent, just a little nudge, and if my attempt at humor caused you any grief at all, I apologize. It's just that it really does help to use Latex in posts as I'm sure you're aware. Thanks for calling this to my attention. Yours, Robert K. "Bob" Lewis – Robert Lewis Oct 13 '13 at 4:54
If you are asked for an expression doe $y'$, then write $y'=\frac{y^3-2xy-12x^2}{x^2-3xy^2}$. But if you are doing further processing, it may be advantageous to work with the "flat" version. – André Nicolas Oct 13 '13 at 4:54
up vote 1 down vote accepted

One cool thing about implicit differentiation of $y$ via an expression such as

$f(x, y) = 0 \tag{1}$

is that the resulting derivative, i.e.,

$\frac{\partial f}{\partial x} + y'(x)\frac{\partial f}{\partial y} = 0, \tag{2}$

is always a linear expression in the function $y'(x)$, and as such, is generally easy to solve for $y'(x)$; indeed, from (2),

$y'(x) = -\frac{\partial f}{\partial x}(\frac{\partial f}{\partial y})^{-1}, \tag{3}$

as long as $\frac{\partial f}{\partial y} \ne 0$. Applying these notions to the case at hand,

$(2xy + x^2 y') - (y^3 + 3xy^2 y') = -12x^2, \tag{4}$

we see it can be re-arranged to give

$(2xy - y^3) + y'(x^2 - 3xy^2) = -12x^2, \tag{5}$

an expression linear in $y'$, for which it is easily solved:

$y'(x) = \frac{y^3 -2xy -12x^2}{x^2 - 3xy^2}, \tag{6}$

which allows $y'(x)$ to be calculated once $x$ and $y$ satisfying $f(x, y) = 0$ are known (here $f(x, y) = 4x^3 + x^2y - xy^3 - 4$).

Hopes this helps. Cheerio,

and as always,

Fiat Lux!!!

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at 5, after you factored out the y', how were you able to rearrange the y^3 and the x^2 if they were in two different groupings? – igknighton Oct 13 '13 at 5:30
It works because $x^2$ is part of the coefficient of $y'$. – Robert Lewis Oct 13 '13 at 5:33

Rather than using $y^\prime$ etc., you might simply want to use the differential operator as $df$, $dx$, $dy$, etc.:

$$d(4x^3 + x^2y - xy^3) = d(4)=0$$

That is,

$$12x^2dx + 2xydx + x^2dy - \cdots =0$$

For example, here $d(x^2y)$ becomes $2xydx + x^2dy $ by the product and chain rules.

Now, group the above as

$$(\operatorname{Expression 1})dx = (\operatorname{Expression 2})dy $$

And from this you get $dy/dx$.

This is a more convenient way to write down calculations of implicit differentiation, once you get the hang of it.

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