Recipe

Differentiate implicitly

Use when: an equation relates x and y but isn't (or can't easily be) solved for y first

Goal: a formula for dy/dx in terms of x and y, valid at any point on the curve

Taught in Implicit differentiation

1. 1

   Apply \dfrac{d}{dx} to every term on both sides of the equation, term by term.

   Why: the equation holds at every point on the curve, so differentiating both sides preserves an equally true statement — you're not solving for y first, just tracking how both sides change together.

2. 2

   Every time you differentiate a term containing y, multiply by \dfrac{dy}{dx} (the chain rule, since y is secretly y(x)).

   Example

   For y^3: \dfrac{d}{dx}[y^3]=3y^2\dfrac{dy}{dx}, not just 3y^2.

   Why: y isn't a free variable here — it's an unknown function of x, and the chain rule requires that inner derivative every time.

3. 3

   A term with both x and y multiplied together (like xy or x^2y^3) needs the product rule, with the dy/dx factor attached to whichever piece it lands on.

   Example

   For xy: \dfrac{d}{dx}[xy]=(1)y+x\left(\dfrac{dy}{dx}\right)=y+x\dfrac{dy}{dx}.

   Why: a mixed term is a product of two functions of x (x itself and y(x)), so it needs the product rule like any other product.

4. 4

   Move every term containing dy/dx to one side of the equation, and everything else to the other.

   Why: dy/dx is what you're solving for — it has to be isolated the same way any unknown does.

5. 5

   Factor dy/dx out of the terms that carry it, then divide to isolate it.

   Example

   For 2x+2y\dfrac{dy}{dx}=0: \dfrac{dy}{dx}=-\dfrac{2x}{2y}=-\dfrac{x}{y}.

   Why: this is ordinary algebra once every dy/dx term is gathered together — the calculus finished at Step 4.