Implicit Differentiation

Differentiating equations where y is not isolated — treating y as a function of x and applying the chain rule to every term.

On a circle equation, implicit differentiation recovers the tangent slope.

Definition

Implicit differentiation finds $\frac{dy}{dx}$ when $y$ is defined implicitly by an equation in $x$ and $y$ , rather than explicitly as $y = f(x)$ .

The key rule: treat $y$ as a function of $x$ , and apply the chain rule whenever you differentiate a term involving $y$ :

$\frac{d}{dx}[y^n] = ny^{n-1} \cdot \frac{dy}{dx}$

Procedure:

Differentiate both sides with respect to $x$ .
Apply the chain rule to every $y$ term (multiply by $\frac{dy}{dx}$ ).
Collect all $\frac{dy}{dx}$ terms on one side.
Solve for $\frac{dy}{dx}$ .

Circle: x² + y² = r²

Differentiate $x^2 + y^2 = r^2$ with respect to $x$ ( $r$ is constant):

$2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$

At the point $(3, 4)$ on a circle of radius $5$ : $\frac{dy}{dx} = -\frac{3}{4}$ .

This matches the geometric fact that the tangent to a circle is perpendicular to the radius (slope of radius $= 4/3$ , perpendicular slope $= -3/4$ ).

Try it

Find $\frac{dy}{dx}$ for the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ . Find the slope at $(0, 2)$ .

Solution

Differentiate: $\frac{2x}{9} + \frac{2y}{4}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{9y}$ .

At $(0, 2)$ : $\frac{dy}{dx} = -\frac{4(0)}{9(2)} = 0$ . The tangent is horizontal — the ellipse is at its top.

Related concepts

Needs first

Derivatives

Related Rates

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