# I am trying to solve the problem: x^2+xy+y^3=0 using implicit differentiation.

I am trying to solve the problem: ${x}^{2}+xy+{y}^{3}=0$ using implicit differentiation.
My workings:
$\left(1\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{d}{dx}\left[{x}^{2}\right]\phantom{\rule{thinmathspace}{0ex}}+\frac{d}{dx}\left[xy\right]\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\frac{d}{dx}\left[{y}^{3}\right]=\frac{d}{dx}\left[0\right]$
$\left(2\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{d{y}^{3}}{dy}\frac{dy}{dx}=0$
$\left(3\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+3{y}^{2}\phantom{\rule{thinmathspace}{0ex}}\left(\frac{dy}{dx}\right)=0$
$\left(4\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}=\overline{)-\frac{2x+y}{3{y}^{2}}}$
But the answer says it should be:
$\left(3\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{dxy}{dy}\frac{dy}{dx}+3{y}^{2}\phantom{\rule{thinmathspace}{0ex}}\left(\frac{dy}{dx}\right)=0$
$\left(4\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{dy}{dx}\left(x+3{y}^{2}\right)=0$
$\left(5\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}=-\frac{2x\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}y}{x\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}3{y}^{2}}$
Why?
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Barbara Klein
You can't conclude that $\frac{d}{dx}\left(xy\right)=y$, since this is effectively the product of the functions $x$ and $y$. Hence, using the product rule gives
$\frac{d}{dx}\left(xy\right)=y\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(y\right)=y+x\frac{dy}{dx}$
The rest should work out as expected.
###### Not exactly what you’re looking for?
Taliyah Reyes
Differentials will make the process much more intuitive.
$\begin{array}{rl}& 0=d\left({x}^{2}+xy+{y}^{3}\right)\\ & 0=2x\phantom{\rule{thinmathspace}{0ex}}dx+\left(x\phantom{\rule{thinmathspace}{0ex}}dy+y\phantom{\rule{thinmathspace}{0ex}}dx\right)+3{y}^{2}dy\\ & \left(x+3{y}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}dy=-\left(2x+y\right)\phantom{\rule{thinmathspace}{0ex}}dx\\ & \frac{dy}{dx}=-\left(\frac{2x+y}{x+3{y}^{2}}\right)\end{array}$