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Here's an exercise from Thomas Calculus, I need to do an implicit differentiation:

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

If I enter this in Wolfram Alpha, I get:

$$ y'(x) =-\frac{3 x^4 + 18 x^3 y + 27 x^2 y^2 - 7 y}{7 x} $$

But if I move things around first like this: $x^3(x+3y)=2x-y$, then I get a completely different result:

$$ y'(x)=-\frac{4x^3+9x^2y-2}{3x^3+1} $$ Which is what I got when doing it by hand. First I thought that it simply took different steps, and one expression can be simplified to another, but then I tried to 3D plot the expressions as functions of $x$ and $y$, and I got different plots. Which makes me wonder how this is possible.

Stanislav Bashkyrtsev
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1 Answers1

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First, the answer provided by Wolfam Alpha in your link includes a negative sign and reads $$y'(x) = - \frac{3x^4 + 18x^3 y + 27 x^2 y^2 - 7y}{7x}. \tag{1}$$ This is a correct expression for the implicit derivative of $y$ with respect to $x$. The second expression $$y'(x) = - \frac{4x^3 + 9x^2 y - 2}{3x^3 + 1} \tag{2}$$ you obtained by hand calculation is also correct. They are not identically equivalent for all $(x,y)$ because they are only equivalent when the original implicit relation $$x^3 = \frac{2x-y}{x+3y} \tag{3}$$ is true; that is to say, $(1)$ and $(2)$ yield the same value of $y'(x)$ for all $(x,y)$ for which $(3)$ is true. To see this, one only needs to explicitly solve $(3)$ for $y$ and perform the differentiation: $$y(x) = \frac{2x - x^4}{1 + 3x^3} \tag{4}$$ hence $$y'(x) = \frac{2 - 16x^3 - 3x^6}{(1 + 3x^3)^2}. \tag{5}$$ Then if we substitute $(4)$ into $(1)$ and simplify, we should obtain $(5)$: $$\begin{align} y'(x) &= -\frac{1}{7x} \left(3x^4 + 18x^3 \frac{2x - x^4}{1 + 3x^3} + 27 x^2 \left(\frac{2x - x^4}{1 + 3x^3}\right)^2 - 7 \frac{2x - x^4}{1 + 3x^3}\right) \\ &= \frac{2 - 16x^3 - 3x^6}{(1+3x^3)^2}. \tag{6} \end{align}$$ The same substitution applied to $(2)$ will also yield $(5)$, the details of which I leave to you as an exercise.

Geometrically, what this means is that if we consider the plots of $(1)$ and $(2)$ as functions $z(x,y)$, they will intersect when $(3)$ is true. This is difficult to see in your figure.

heropup
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  • Takes me awhile, but I'm starting to understand a little. Still figuring some nuances, but in the meantime.. Is there a simple strategy then to check the results? It's not always possible to express $y$ using just $x$. Like if we had $x^3=\frac{2x-y^2}{x+3y}$ instead - it doesn't seem possible to express $y$ without referencing $y$ itself. – Stanislav Bashkyrtsev Mar 26 '24 at 15:55