Implicitly differentiate $x^3 + y^3 = 3xy$

Question

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?

Answer 1

Implicit differentiation is really quite simple. Think of $x$ as the independent variable, and think of $y$ as a function of $x$.

So, if you take the derivative of $x^9$ with respect to $x$, you get $9x^8$. But if you take the derivative of $y^9$ with respect to $x$, you get $9y^8\cdot y'$.

Why? It's exactly like taking the derivative of $[f(x)]^9$, so use the chain rule to get $9[f(x)]^8\cdot f'(x)$.

In an equation, some terms may contain $y$ and some may not, so you will typically find $y'$ scattered here and there on both sides after you differentiate implicitly. If your goal is to find the derivative, you will have to solve the resulting equation for $y'$, so your answer will usually involve both $x$ and $y$.

Answer 2

Think of $y$ as a function of $x$. I will explicitly write this as $y(x)$. Then proceed to differentiate everything with respect to $x$ as normal, remembering the chain rule.

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

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

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