Definition
An implicit function is a function where the dependent variable $y$ is not isolated on one side of the equation. Instead, the relationship between $x$ and $y$ is expressed as:
$R(x, y) = 0$
In other words, the output $y$ is "implied" by the equation rather than being explicitly defined in terms of $input$ $x$.
| Type | Form | Example |
|---|---|---|
| Explicit | $y = f(x)$ | $y = \sqrt{4 - x^2}$ |
| Implicit | $R(x, y) = 0$ | $x^2 + y^2 = 4$ |
Why use Implicit?
* Some curves (like circles or loops) cannot be written as a single $y = f(x)$ function.
* It simplifies complex algebraic relations where solving for $y$ is difficult or impossible.
if $R(x, y) = 0$, the derivative of the implicit function $y(x)$ is given by
$\frac{dy}{dx} = -\frac{\frac{\partial R}{\partial x}}{\frac{\partial R}{\partial y}} = -\frac{R_x}{R_y}$
where $R_x$ and $R_y$ indicate the partial derivatives of $R$ with respect to $x$ and $y$.
Consider an implicit equation $R(x, y) = 0$, where $y$ is a differentiable function of $x$. We can use the Multivariable Chain Rule to differentiate both sides with respect to $x$:
$$\frac{d}{dx}[R(x, y)] = \frac{d}{dx}[0]$$
$$\frac{\partial R}{\partial x} \cdot \frac{dx}{dx} + \frac{\partial R}{\partial y} \cdot \frac{dy}{dx} = 0$$
Since $\frac{dx}{dx} = 1$, the equation simplifies to:
$$R_x + R_y \cdot \frac{dy}{dx} = 0$$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = -\frac{R_x}{R_y}$$
Q.E.D. — This confirms the relationship between partial derivatives and the implicit derivative.
Find $\frac{dy}{dx}$ for the curve: $x^2 + y^2 = 4$
Step 1: Define $R(x, y)$
Let $R(x, y) = x^2 + y^2 - 4 = 0$
Step 2: Partial Derivatives
$R_x = \frac{\partial}{\partial x}(x^2 + y^2 - 4) = 2x$
$R_y = \frac{\partial}{\partial y}(x^2 + y^2 - 4) = 2y$
Step 3: Apply the Formula
$$\frac{dy}{dx} = -\frac{R_x}{R_y} = -\frac{2x}{2y} = -\frac{x}{y}$$
Note: The slope is undefined when $y=0$ (vertical tangents).
Find $\frac{dy}{dx}$ for the curve: $x^3 + y^3 = 6xy$
Step 1: Set the implicit form $R(x, y)$
Move all terms to one side: $R(x, y) = x^3 + y^3 - 6xy = 0$
Step 2: Partial Derivatives
Treat $y$ as constant: $R_x = \frac{\partial}{\partial x}(x^3 + y^3 - 6xy) = 3x^2 - 6y$
Treat $x$ as constant: $R_y = \frac{\partial}{\partial y}(x^3 + y^3 - 6xy) = 3y^2 - 6x$
Step 3: Solve for $\frac{dy}{dx}$
$$\frac{dy}{dx} = -\frac{R_x}{R_y} = -\frac{3x^2 - 6y}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$$
This curve forms a loop in the first quadrant and is a classic example of implicit relations.