# Use implicit differentiation to find dy/dx for the equation x^3+y^3=3xy.

I have this practice problem before a test. Use implicit differentiation to find $dy/dx$ for the equation
${x}^{3}+{y}^{3}=3xy.$
I have no idea how to do this, I didn't understand my lecturer. Can you guys show me the steps?
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Caylee Davenport
Think of $y$ as a function of $x$. I will explicitly write this as $y\left(x\right)$. Then proceed to differentiate everything with respect to $x$ as normal, remembering the chain rule.

We need to differentiate ${x}^{3}+{y}^{3}\left(x\right)=3xy\left(x\right)$. Let's do each term one by one.

1. Differentiate ${x}^{3}$. You should quickly see this is $3{x}^{2}$.
2. To differentiate $\left(y\left(x\right){\right)}^{3}$, we need to remember the chain rule. This can be written in many different ways, but this is a composition of the functions $\left(\cdot {\right)}^{3}\circ y\circ x$. The derivative is $3\left(y\left(x\right){\right)}^{2}{y}^{\prime }\left(x\right)$
3. To differentiate $3xy\left(x\right)$, you must remember the product rule. The derivative is $3y\ast \left(x\right)+3x{y}^{\prime }\left(x\right)$.

So in total, differentiation yields
$3{x}^{2}+3{y}^{2}\left(x\right){y}^{\prime }\left(x\right)=3y\left(x\right)+3x{y}^{\prime }\left(x\right).$
In this form, to find ${y}^{\prime }$, you isolate it (if possible) in this equation. Here, this simplifies (after cancelling factors of 3) to
${y}^{\prime }\left(x\right)=\frac{y\left(x\right)-{x}^{2}}{{y}^{2}\left(x\right)-x}.$