Implicit Differentiation (Multivariable)
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.5: “The Chain Rule” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-5-the-chain-rule |
| Textbook used in class | Stewart, Calculus, Section 14.5: “The Chain Rule” (Examples 4, 5) |
Quick Reference
Curve $F(x,y) = 0$ (implicit function of $x$): $$\frac{dy}{dx} = -\frac{F_x}{F_y} \quad \text{(when } F_y \neq 0\text{)}.$$
Surface $F(x,y,z) = 0$ (implicit function): $$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} \quad \text{(when } F_z \neq 0\text{)}.$$
Motivation
When a curve is defined implicitly by $F(x,y) = 0$ (rather than explicitly as $y = f(x)$), single-variable implicit differentiation requires differentiating both sides with respect to $x$ and solving for $dy/dx$. The multivariable chain rule provides a systematic formula that skips the algebra: $dy/dx = -F_x/F_y$.
The same idea extends to surfaces $F(x,y,z) = 0$, giving partial derivatives without ever solving for $z$ explicitly.
Key Concepts
1. Deriving the Formula
Suppose $F(x,y) = 0$ defines $y$ implicitly as a function of $x$. Differentiate both sides with respect to $x$, treating $y = y(x)$:
$$\frac{d}{dx}F(x,y(x)) = 0.$$
By the chain rule: $F_x \cdot 1 + F_y \cdot \dfrac{dy}{dx} = 0$.
Solving: $\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$.
2. Surface Case
For $F(x,y,z) = 0$ defining $z = z(x,y)$ implicitly, differentiate with respect to $x$ (holding $y$ fixed):
$$F_x + F_z\frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}.$$
This requires only computing partial derivatives of $F$, not solving for $z$.
Worked Examples
Example 1. Find $dy/dx$ for $x^3 + y^3 = 6xy$ (the folium of Descartes).
Define $F(x,y) = x^3 + y^3 - 6xy$.
$F_x = 3x^2 - 6y$, $F_y = 3y^2 - 6x$.
$$\frac{dy}{dx} = -\frac{3x^2 - 6y}{3y^2 - 6x} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.$$
Example 2. Find $\partial z/\partial x$ and $\partial z/\partial y$ for $x^2 + y^2 + z^2 = 1$ (the unit sphere).
$F(x,y,z) = x^2 + y^2 + z^2 - 1$, so $F_x = 2x$, $F_y = 2y$, $F_z = 2z$.
$$\frac{\partial z}{\partial x} = -\frac{2x}{2z} = -\frac{x}{z}, \qquad \frac{\partial z}{\partial y} = -\frac{y}{z}.$$
(Valid where $z \neq 0$, i.e., away from the equator of the sphere.)
to find $\partial z/\partial x$ from $F(x,y,z) = 0$, set $F_y = 0$ and then solve. The formula is $\partial z/\partial x = -F_x/F_z$, not $-F_x/F_y$. The denominator involves the partial derivative with respect to the variable being solved for ($z$), not with respect to $y$. Similarly, $\partial z/\partial y = -F_y/F_z$. A common mix-up is using $F_y$ in the denominator.
Leveled Practice
Problem 1. Find $dy/dx$ for $\cos(xy) + y\ln x = 0$.
Show answer
$F(x,y) = \cos(xy) + y\ln x$.
$F_x = -y\sin(xy) + y/x$.
$F_y = -x\sin(xy) + \ln x$.
$dy/dx = -\dfrac{-y\sin(xy)+y/x}{-x\sin(xy)+\ln x} = \dfrac{y\sin(xy)-y/x}{-x\sin(xy)+\ln x}$.