# Y^3+XY+X^2=0 and I was suppose to find dY/dX and substitute the value 10 for X after differentiating. Weirdly enough, this equation does not let me get rid of the variable Y after solving and hence I couldn't find a valid answer. Would request you to help me approach the problem. Also, I would like to do so without finding the roots of Y and solving, as someone pointed out earlier in class.

${Y}^{3}+XY+{X}^{2}=0$
and I was suppose to find dY/dX and substitute the value 10 for X after differentiating. Weirdly enough, this equation does not let me get rid of the variable Y after solving and hence I couldn't find a valid answer.

Would request you to help me approach the problem. Also, I would like to do so without finding the roots of Y and solving, as someone pointed out earlier in class.
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xgirlrogueim
Let $F={y}^{3}+xy+{x}^{2}$. Then
${F}_{x}^{\prime }=y+2x$
${F}_{y}^{\prime }=3{y}^{2}+x$
$\frac{dy}{dx}=-\frac{{F}_{x}^{\prime }}{{F}_{y}^{\prime }}=-\frac{y+2x}{3{y}^{2}+x}$
We know $x=10$ and we need the corresponding y which is the solution of ${y}^{3}+10y+100=0$ which is cubic equation that you can solve using Cardano. This equation has a single real root given by
$y=\frac{\sqrt[3]{10\left(\sqrt{2055}-45\right)}}{{3}^{2/3}}-\frac{{10}^{2/3}}{\sqrt[3]{3\left(\sqrt{2055}-45\right)}}\simeq -3.93003$
If you do not use Cardano, only a numerical method could be used. The simplest would be Newton, which, starting from a "reason able" guess ${y}_{0}$ will update it according to
${y}_{n+1}={y}_{n}-\frac{f\left({y}_{n}\right)}{{f}^{\prime }\left({y}_{n}\right)}$
with $f\left(y\right)={y}^{3}+10y+100$. By inspection, you could see that there is a root close to −4; so, let us select ${y}_{0}=-4$. Soo, Newton iterates will be −3.93103,−3.93003.
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planhetkk
${Y}^{3}+XY+{X}^{2}=0$
$3{Y}^{2}dY+YdX+XdY+2XdX=0$
$\left(3{Y}^{2}+X\right)dY=-\left(Y+2X\right)dX$
$\frac{dY}{dX}=-\frac{Y+2X}{3{Y}^{2}+X}$
Are you sure that you need more ? If yes, you will have to solve ${Y}^{3}+XY+{X}^{2}=0$ for $Y$ and put it back into the above formula.