Implicit Differentiation Formula

The Formula

For F(x, y) = 0: differentiate both sides with respect to x, apply chain rule to y-terms, solve for \frac{dy}{dx}.

When to use: Sometimes you can't (or don't want to) solve for y explicitly. Instead, differentiate the whole equation as-is. Every time you differentiate a y-term, attach \frac{dy}{dx} by the chain rule (since y secretly depends on x), then solve for \frac{dy}{dx}.

Quick Example

Find \frac{dy}{dx} for the circle x^2 + y^2 = 25.
Differentiate: 2x + 2y\frac{dy}{dx} = 0.
Solve: \frac{dy}{dx} = -\frac{x}{y}
At (3, 4): slope = -\frac{3}{4}.

Notation

\frac{dy}{dx} found implicitly. Alternatively, \frac{dy}{dx} = -\frac{F_x}{F_y} where F_x and F_y are partial derivatives of F(x,y).

What This Formula Means

A technique for finding \frac{dy}{dx} when y is defined implicitly by an equation F(x, y) = 0 rather than explicitly as y = f(x). Differentiate both sides with respect to x, treating y as a function of x, then solve for \frac{dy}{dx}.

Sometimes you can't (or don't want to) solve for y explicitly. Instead, differentiate the whole equation as-is. Every time you differentiate a y-term, attach \frac{dy}{dx} by the chain rule (since y secretly depends on x), then solve for \frac{dy}{dx}.

Formal View

If F(x, y) = 0 defines y implicitly as a differentiable function of x, then by the chain rule: \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} = 0, so \frac{dy}{dx} = -\frac{F_x(x,y)}{F_y(x,y)} provided F_y(x,y) \neq 0.

Worked Examples

Example 1

easy
Find \frac{dy}{dx} for the circle x^2 + y^2 = 25 and evaluate it at the point (3, 4).

Solution

  1. 1
    Differentiate both sides with respect to x: 2x + 2y\frac{dy}{dx} = 0.
  2. 2
    Solve for \frac{dy}{dx}: \frac{dy}{dx} = -\frac{x}{y}.
  3. 3
    At (3, 4): \frac{dy}{dx} = -\frac{3}{4}.

Answer

\dfrac{dy}{dx} = -\dfrac{x}{y}; at (3, 4): slope = -\dfrac{3}{4}
Whenever a y-term is differentiated, attach \frac{dy}{dx} by the chain rule. Then collect all \frac{dy}{dx} terms and solve. The tangent to a circle at (3,4) has slope -3/4.

Example 2

hard
Find \frac{dy}{dx} for x^3 + y^3 = 6xy (folium of Descartes).

Common Mistakes

  • Forgetting the \frac{dy}{dx} factor when differentiating y-terms: \frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}, NOT 3y^2.
  • Not using the product rule when x and y are multiplied: \frac{d}{dx}[xy] = x\frac{dy}{dx} + y, NOT just x\frac{dy}{dx} or just y.
  • Getting confused about when the answer contains y: implicit derivatives typically have both x and y in the result, which is fine—you need a specific point (x, y) on the curve to get a numerical slope.

Why This Formula Matters

Many important curves (circles, ellipses, hyperbolas) and equations in physics can't be easily solved for y. Implicit differentiation lets you find slopes, tangent lines, and rates of change without solving for y first.

Frequently Asked Questions

What is the Implicit Differentiation formula?

A technique for finding \frac{dy}{dx} when y is defined implicitly by an equation F(x, y) = 0 rather than explicitly as y = f(x). Differentiate both sides with respect to x, treating y as a function of x, then solve for \frac{dy}{dx}.

How do you use the Implicit Differentiation formula?

Sometimes you can't (or don't want to) solve for y explicitly. Instead, differentiate the whole equation as-is. Every time you differentiate a y-term, attach \frac{dy}{dx} by the chain rule (since y secretly depends on x), then solve for \frac{dy}{dx}.

What do the symbols mean in the Implicit Differentiation formula?

\frac{dy}{dx} found implicitly. Alternatively, \frac{dy}{dx} = -\frac{F_x}{F_y} where F_x and F_y are partial derivatives of F(x,y).

Why is the Implicit Differentiation formula important in Math?

Many important curves (circles, ellipses, hyperbolas) and equations in physics can't be easily solved for y. Implicit differentiation lets you find slopes, tangent lines, and rates of change without solving for y first.

What do students get wrong about Implicit Differentiation?

Don't forget \frac{dy}{dx} every time you differentiate a term involving y. For example, \frac{d}{dx}[xy] = x\frac{dy}{dx} + y by the product rule, with \frac{dy}{dx} appearing because y depends on x.

What should I learn before the Implicit Differentiation formula?

Before studying the Implicit Differentiation formula, you should understand: derivative, chain rule, differentiation rules.

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Derivatives Explained: Rules, Interpretation, and Applications →
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